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Position of point wrt ellipse - class-XII

Description: position of point wrt ellipse
Number of Questions: 64
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Tags: maths ellipse
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The dist.of a point P on the ellipse $\cfrac{{{x^2}}}{{12}} + \cfrac{{{y^2}}}{4} = 1$ from centre is $\sqrt 6 $ then the eccentric angle of P is 

  1. $\cfrac{\pi }{2}$

  2. $\cfrac{\pi }{6}$

  3. $\cfrac{\pi }{4}$

  4. $\cfrac{\pi }{3}$


Correct Option: D
Explanation:
Let the point be $P = (a\cos\theta , b\sin\theta )$
 
Distance of point $P$ from centre is $\sqrt 6$
$\therefore \sqrt { { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta} = \sqrt { 6 }$
$\Rightarrow { a }^{ 2 }\cos^{ 2 }\theta + b^{ 2 }\sin^{ 2 }\theta = 6$
$\Rightarrow 12\cos^{ 2 }\theta + 4\sin^{ 2 }\theta = 6 \cos^{ 2 }\theta + \sin^{ 2 }\theta = 1$
$\Rightarrow 8\cos^{ 2 }\theta = 2$
$\Rightarrow \cos\theta = \cfrac { 1 }{ 2 }$
Hence, $\theta = \cfrac { \pi }{ 3 }$

Point $(1,2)$ lies _____ the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.

  1. inside

  2. outside

  3. on

  4. None of the above


Correct Option: A
Explanation:

The region (disk) bounded by the ellipse is given by the equation:

$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}\leq 1$ centered at $(h,k)$ ..... $(1)$
Point $(x,y)$ lies inside the ellipse if it satisfies $(1)$
Take $(x,y)=(1,2)$
Consider, $\dfrac{(x-0)^2}{16}+\dfrac{(y-0)^2}{9}$

                 $=\dfrac{1^{2}}{16}+\dfrac{2^{2}}{9}=\dfrac{73}{144}<1$
Hence, $(1,2)$ lies inside the given ellipse.

Eccentric angle of a point on the ellipse $x^{2}+3y^{2}=6$ at a distance $2$ units. from the centre of the ellipse is

  1. $2\pi/3$

  2. $\pi/3$

  3. $4\pi/3$

  4. $none\ of\ these$


Correct Option: B

Let the equation of the ellipse be $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Let $f(x,y) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$. To determine whether the point $(x _1,y _1)$ lies inside the ellipse, the necessary condition is:

  1. $f(x _1,y _1) < 0$

  2. $f(x _1,y _1) > 0$

  3. $f(x _1,y _1) = 0$

  4. None of these


Correct Option: A
Explanation:

$f(x,y) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$ ........ $(1)

The region (disk) bounded by the ellipse is given by the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}\leq 1$ centered at $(h,k)$.

The given equation of ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$ centered at origin i.e. $(0,0)$.
The region bounded by this ellipse is 
$\dfrac{(x)^2}{a^2}+\dfrac{(y)^2}{b^2}\leq 1$ ...... $(1)$
The point $x _{1},y _{1}$ lies inside the given ellipse if it satisfies $(1)$
i.e. if $\dfrac{(x _{1})^2}{a^2}+\dfrac{(y _{1})^2}{b^2}\leq 1$ ...... $(1)$
if $\dfrac{(x _{1})^2}{a^2}+\dfrac{(y _{1})^2}{b^2} - 1<0$ ...... $(1)$
$\implies$ $f(x _{1}, y _{1})<0$  ....... From $(1)$
Hence, option A is correct.

The locus of a point whose distance form the point $(3,0)$ is $3/5$ times its distance from the line $x=p$ is an ellipse with centre at the origin. The value of $p$ is 

  1. $5$

  2. $7$

  3. $\dfrac{25}{3}$

  4. $\dfrac{25}{9}$


Correct Option: C

Point $\left(\sqrt5, \dfrac4{\sqrt5}\right)$ lies _____ the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{4} = 1$.

  1. inside

  2. outside

  3. on

  4. None of the above


Correct Option: C
Explanation:

The point $(x,y)$ lies inside the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if it satisfies the following condition i.e.

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\leq 1$
To check $\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$ lies in the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{4} = 1$
Here $a=2$ and $b=5$
Take $(x,y)=\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$
$\therefore$ $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{(\sqrt5)^2}{4}+\dfrac{(\frac{4}{\sqrt5})^2}{25}=\dfrac{5}{4}+\dfrac{16}{5\times 25}=\dfrac{689}{600} < 1$
Hence, $\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$ lies in the given ellipse.

State the following statement is true or false
$(3,0),(0,4)$ and $(4,0)$ are all points that lie on the ellipse.

  1. True

  2. False


Correct Option: B

Determine position of a point $(2,3)$ with respect to the ellipse $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{25}=1$.

  1. Outside

  2. Inside

  3. On the ellipse

  4. None of the above


Correct Option: B
Explanation:

Given equation is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1$

$ \dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1=0$

The curve is defined by, 

$f(x,y)=\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { { (2) }^{ 2 } }{ 16 } +\dfrac { { (3) }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { 1 }{ 4 } +\dfrac { 9 }{ 25 } -1=\dfrac { 25+36-100 }{ 100 } =-\dfrac { 39 }{ 100 } \\ f(2,3)<0$

So, the point lies inside the ellipse.

Option B is correct.

Position of a point $(3,-4))$ with respect to the ellipse $16x^{2}+9y^{2}=144$ lies 

  1. outside

  2. inside

  3. on

  4. None of the above


Correct Option: A
Explanation:

Given equation is $16{ x }^{ 2 }+9{ y }^{ 2 }=144$

$ f(x,y)=16{ x }^{ 2 }+9{ y }^{ 2 }-144\ f(3,-4)=16{ (3) }^{ 2 }+9(-4)^{ 2 }-144\ f(3,-4)=144+144-144\ f(3,-4)=144\ f(3,-4)>0$
So, the point lies outside the ellipse.

Option A is correct.

The position of point $(4,3)$ with respect to the ellipse $\dfrac{x^2}{4}+\dfrac {y^2}{3}=1$

  1. Inside 

  2. Outside

  3. On the Ellipse 

  4. None.


Correct Option: B
Explanation:

Given ellipse $\dfrac {x^2}{4}+\dfrac {y^2}{3}=1$


Given point $(4,3)$

The position of point is $\dfrac {4^2}{4}+\dfrac {3^2}3-1\4+3-1=6>0$
So the point s outside the ellipse 

The distance of a point P on the ellipse $\dfrac{{{x^2}}}{{12}} + \dfrac{{{y^2}}}{4} = 1$ from centre is $\sqrt 6 $ then the ecentric angle of P is

  1. $\dfrac{2\pi }{3}$

  2. $\dfrac{\pi }{6}$

  3. $\dfrac{\pi }{4}$

  4. $\dfrac{\pi }{3}$


Correct Option: A,D
Explanation:

Ellipse$:\cfrac { x^{ 2 } }{ 12 } +\cfrac { y^{ 2 } }{ 4 } =1$

Centre$:C(0,0)$
Given that $PC=\sqrt { 6 } $
Parametric coordinates $:y=b\sin { \phi  } ,x=a\cos { \phi  } $
$\phi$ is eccentric angle. 
$\Rightarrow 6=(2\sqrt { 3 } \cos { \phi  } -0)^{ 2 }+(2\sin { \phi  } -0)^{ 2 }$
$ \Rightarrow 6=12\cos { ^{ 2 }\phi  } +4\sin { ^{ 2 }\phi  } $
$ \Rightarrow 2=8\cos ^{ 2 }{ \phi  } $
$ \Rightarrow \cos ^{ 2 }{ \phi  } =\cfrac { 1 }{ 4 } $
$ \Rightarrow \cos { \phi  } = \pm \cfrac { 1 }{ 2 } $
$ \Rightarrow \phi = \cfrac { \pi  }{ 3 } $
$\Rightarrow \phi =\cfrac { 2\pi  }{ 3 } $

Evaluate $\displaystyle \int x^2+3x+5\ dx$ 

  1. $\dfrac {x^3}3+3\dfrac {x^2}3+\dfrac 53+c$

  2. $\dfrac {x^2}2+ 3x+5+c $

  3. $ \dfrac {x^3}3+3\dfrac {x^2}2+5x+c $

  4. None of the above.


Correct Option: C
Explanation:

$\displaystyle \int x^2+3x+5\ dx$ 


$=\displaystyle \int x^2 dx+\int 3x dx+\int 5 dx$


$=\dfrac {x^3}{3}+\dfrac {3x^2}{2}+5x+C$


An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is $\dfrac 23$ then the eccentricity of the ellipse is  

  1. $\dfrac{2\sqrt{2}}{3}$

  2. $\sqrt{5}$

  3. $8$

  4. $2$


Correct Option: A
Explanation:
Probability  $p=\dfrac 23$

Let the radius of circle$=a$

Major axis$=2a$

Minor axis$=2b$

Area of circles$=\pi a^2$

Area of ellipse$=\pi ab$

as because

$P=\dfrac{\pi ab-\pi a^2}{\pi ab}=1-\dfrac{a}{b}$

$\dfrac{a}{b}=1-\dfrac{2}{3}=\dfrac{1}{3}$

$e=\sqrt{1-\left(\dfrac{a}{b}\right)^2}-\sqrt{1-\left(\dfrac{1}{3}\right)^2}$

$=\sqrt{\dfrac{8}{9}}$

Eccentricity$=e=\dfrac{2\sqrt{2}}{3}$.

The position of the point (1,3)n with respect to the ellipse $4x^{2}+9y^{2}-16x-54y+61=0$ is 

  1. Outside the ellipse 

  2. On the ellipse

  3. On the major axis

  4. On the minor axis


Correct Option: A
Explanation:

We have,

Equation of given ellipse,

$4{{x}^{2}}+9{{y}^{2}}-16x-54y+61=0$

Given point,

Let, $\left( x,y \right)=\left( 1,\,3 \right)$

Then, position of point in given ellipse is

$ 4{{\left( 1 \right)}^{2}}+9{{\left( 3 \right)}^{2}}-16\left( 1 \right)-54\left( 3 \right)+61=0 $

$ \Rightarrow 4+81-16-162+61=0 $

$ \Rightarrow 146-178=0 $

$ \Rightarrow -32<0 $

The position of point outside the ellipse.

If the point $(a\sin\theta, a\cos\theta)$ lies on the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ then the value of $\sin 2\theta$ is (where $a\neq b, a>0, b>0$ and $e$ is the eccentricity of the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$)

  1. $\dfrac{2\sqrt{1+e^{2}}}{2+e^{2}}$

  2. $\dfrac{2\sqrt{1-e^{2}}}{2+e^{2}}$

  3. $\dfrac{2\sqrt{1-e^{2}}}{2-e^{2}}$

  4. $\dfrac{2\sqrt{1+e^{2}}} {2-e^{2}}$


Correct Option: A

The locus of a point whose chord of contact to the ellipse $x^{2}+2y^{2}=1$ subtends a right angle at the centre of the ellipese is 

  1. $x^{2}+4y^{2}=3$

  2. $y^{2}=4x$

  3. $2x^{2}+y^{2}=1$

  4. none of these


Correct Option: A
Explanation:

Equation of ellipse-

${x}^{2} + 2 {y}^{2} = 1 ..... \; \left(  1 \right)$
Let $\left( h, k \right)$ be the point whose chord of contact subtends a right angle at the centre of ellipse.
Equation of chord of contact-
$hx + 2ky = 1$
Squaring both sides, we have
${\left( hx + 2ky \right)}^{2} = {\left( 1 \right)}^{2}$
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = 1 ..... \left( 2 \right)$
Now, from equation $\left( 1 \right) &amp; \left( 2 \right)$, we have
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = {x}^{2} + 2 {y}^{2}$
$\Rightarrow \left( {h}^{2} - 1 \right) {x}^{2} + \left( 4 {k}^{2} - 2 \right) {y}^{2} + 4hkxy = 0$
The above equation represents a pair of perpendicular lines if
Coefficient of ${x}^{2} + $ Coefficient of ${y}^{2} = 0$
$\left( {h}^{2} - 1 \right) + \left( 4 {k}^{2} - 2 \right) = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} - 3 = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} = 3$
Replacing $h$ and $k$ with $x$ and $y$ respectively, we get
${x}^{2} + 4 {y}^{2} = 3$
Hence the locus of the point whose chord of contact subtends a right angle at the centre of ellipse is ${x}^{2} + 4 {y}^{2} = 3$.

Equation of the largest circle with centre (1,0) that can be inscribed in the ellipse $x^2 + 4y^2 = 16$ is 

  1. $2x^2 + 2y^2 - 4x + 7 = 0$

  2. $x^2 + y^2 - 2x + 5 = 0$

  3. $3x^2 + 3y^2 - 6x - 8 = 0$

  4. None of these


Correct Option: C
Explanation:
$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{4}=1$

Point on the ellipse $(4\cos\theta, 2\sin\theta)$

Let the circle have radius $=r$

$(x-1)^{2}+(y-0)^{2}=r^{2}$ Solving if with ellipse

$x^{2}+4y^{2}=16$

$(x-1)^{2}+\dfrac{(16-x^{2})}{4}=r^{2}$

$4(x^{2}-2x+1)+16-x^{2}=4r^{2}$

$3x^{2}-8x+20-4r^{2}=0$

As the circle & ellipse touch each other 

$D=0$

$8^{2}-4.2\times (20-4r^{2})=0$

$r^{2}=\dfrac{\pi}{3}$

$(x-1)^{2}+y^{2}=\dfrac{11}{3}$

$3x^{2}+3y^{2}-6x-8=0$

An ellipse of major axis $20\sqrt {3}$ and minor axis $20$ slides along the coordinate axes and always remains confined in the $1^{st}$ quadrant. The locus of the centre of the ellipse therefore describes the arc of a circle. The length of this arc is

  1. $5\pi$

  2. $20\pi$

  3. $\dfrac {5\pi}{3}$

  4. $\dfrac {20\pi}{3}$


Correct Option: B

A tangent to the ellipse $4x^2+9y^2=36$ is cut by tangent at the extremities of the major axis at $T$ and $T'$. The circles on $TT'$ as diameters passes through the point 

  1. $(0,-\sqrt5)$

  2. $(\sqrt5,0)$

  3. $(0,0)$

  4. $(3,2)$


Correct Option: B

If the line $x\, cos\, \alpha+y\,sin \,\alpha=p$ is normal to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, then 

  1. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=a^2-b^2$

  2. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=(a^2-b^2)^2$

  3. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=a^2-b^2$

  4. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=(a^2-b^2)^2$


Correct Option: A

Let $(a, 0)$ and $B(b, 0)$ be fixed distinct points on the $x-axis$, none of which coincides with the origin $O(0, 0)$ and let $C$ be a point on the $y-axis$. Let $L$ be a line through the $O(0, 0)$ and perpendicular to the line $AC$, The locus of the point of intersection of lines $L$ and $BC$ if $C$ varies along the $y-axis$, is (provided $x^{2}+ab\neq 0$) 

  1. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=x$

  2. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=y$

  3. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=x$

  4. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=y$


Correct Option: D

If P($\theta$) and Q($\pi$/2 + $\theta$) are two points on the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Locus of the mid-point of PQ is

  1. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{2}$

  2. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 4$

  3. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2$

  4. None of these


Correct Option: A
Explanation:

Given,$P(\theta),Q(\frac{\pi}{2}+\theta)$ are two points on the ellipse $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$
Any point on the ellipse will be $(acos\theta,bsin\theta)$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(acos(\frac{\pi}{2}+\theta),bsin(\frac{\pi}{2}+\theta))$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(-asin\theta,bcos\theta)$
Let required point be $C(x,y)$
Given, $C=mid-point\;of\;PQ$
$\Rightarrow (x,y)=(\displaystyle\frac{(acos\theta-asin\theta)}{2},\displaystyle\frac{(bsin\theta+bcos\theta)}{2})$
$\Rightarrow \displaystyle\frac{x}{a}=(\displaystyle\frac{(cos\theta-sin\theta)}{2}),\displaystyle\frac{y}{b}=(\displaystyle\frac{(sin\theta+cos\theta)}{2})$
on squaring $\displaystyle\frac{x}{a},\displaystyle\frac{y}{b}$ and adding both
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=(\displaystyle\frac{(cos\theta-sin\theta)}{2})^2+(\displaystyle\frac{(sin\theta+cos\theta)}{2})^2$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{2(cos^2\theta+sin^2\theta)}{4}$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{1}{2}$(since $cos^2\theta+sin^2\theta=1$)

The value of $\alpha$ for which the point $(\alpha,\alpha+2)$ is an interior point of smaller segment of the curve $x^{2}+y^{2}-4=0$ made by the chord of the curve whose equation is $3x+4y+12=0$ is

  1. $\left(-\infty,\dfrac {-20}{7}\right)$

  2. $(-2,0)$

  3. $\left(-\infty,\dfrac {20}{7}\right)$

  4. $\alpha\ \epsilon\ \phi$


Correct Option: A

The distance of a point on the ellipse $\dfrac {x^{2}}{6}+\dfrac {y^{2}}{2}=1$ from the centre is $2$, then the eccentric angle is-

  1. $\dfrac \pi3$

  2. $\dfrac \pi4$

  3. $\dfrac \pi6$

  4. $\dfrac \pi2$


Correct Option: B
Explanation:
Given ellipse $ \dfrac{x^{2}}{6}+\dfrac{y^{2}}{2} = 1 $

Let $ \theta$ be the eccentric angle of the point $p$

coordinate of $p$ $ (\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$

Given distance $= 2 $

$ \therefore $ $OP = 2$

$ \sqrt{6 cos^{2}\theta +2sin^{2}\theta } = 2 \Rightarrow 6cos^{2}\theta +2sin^{2}\theta  = 4$

$ 3cos^{2}\theta +sin^{2}\theta  = 2 $

$ 2 sin^{2}\theta  = 1$

$ sin^{2}\theta  = \dfrac{1}{2} \Rightarrow  sin\theta  = \pm  \dfrac{1}{\sqrt{2}}$

$ \therefore $ eccentric angle $\theta  = \pm \dfrac{\pi }{4}$

A rod of length $l$ rests against a vertical wall and a floor of a room.Let P be a point on the rod,nearer to its end on the wall, that divides its length in the ratio 1:2 if the rod begins to slide on the floor,then the locus of P is:

  1. an ellipse of eccentricity $\dfrac { 1 }{ 2 }$

  2. an ellipse of eccentricity $\dfrac { \sqrt { 3 } }{ 2 }$

  3. a circle of radius $\dfrac { l }{ 2 }$

  4. a circle of radius $\dfrac { \sqrt { 3 } }{ 2 } l$


Correct Option: B

The distance from the foci of $P(a,b)$ on the ellipse $\dfrac {x^{2}}{9}+\dfrac {y^{2}}{25}=1$ are

  1. $4\pm \dfrac {5}{4}b$

  2. $5\pm \dfrac {4}{5}a$

  3. $5\pm \dfrac {4}{5}b$

  4. $none\ of\ these$


Correct Option: C

The number of rational points on the ellipse $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1$ is

  1. $\infty$

  2. $4$

  3. $0$

  4. $2$


Correct Option: A

In an ellipse the distance between its foci is 6 and its minor axis is 8 . Its eccentricity is

  1. $\dfrac{6}{5}$

  2. $\dfrac{4}{5}$

  3. $\dfrac{3}{5}$

  4. $\dfrac{3}{2}$


Correct Option: C
Explanation:
Given that $2ae=6$ and $2b=8$

$\Rightarrow ae=3$      $\Rightarrow b=4$

$b^2=a^2(1-e^2)$

$b^2=a^2-a^2e^2$

$16=a^2-9$

$\Rightarrow a^2=25$

$\Rightarrow a=5$

$5e=3$

$\Rightarrow e=\dfrac{3}{5}$.

A point on the ellipse is $\displaystyle \frac{x^{2}}{6} + \frac{y^{2}}{2} = 1$ at a distance of $2$ from the centre of the ellipse has the eccentric angle

  1. $\displaystyle \frac{\pi}{4}$

  2. $\displaystyle \frac{\pi}{3}$

  3. $\displaystyle \frac{\pi}{6}$

  4. $\displaystyle \frac{\pi}{2}$


Correct Option: A
Explanation:

Given, equation of ellipse as $\displaystyle\frac{x^2}{6}+\displaystyle\frac{y^2}{2}=1$(where length of major axis=$\sqrt6$,length of minor axis=$\sqrt2$)
center of ellipse is (0,0) which is parallel to horizontal axis and with eccentricity 'e'.
Any point on the ellipse will be as $(acos\theta,bsin\theta)\Rightarrow P(\sqrt6cos\theta,\sqrt2sin\theta)$
Distance of point P from center=2
$\Rightarrow \sqrt((\sqrt6cos\theta-0)^2+(\sqrt2sin\theta-0)^2)=2$
$\Rightarrow (6cos^2\theta+2sin^2\theta)=4$
$\Rightarrow (3cos^2\theta+sin^2\theta)=2$
$\Rightarrow 2cos^2\theta+1=2$
$\Rightarrow cos\theta=\pm\frac{1}{\sqrt2}$
$\Rightarrow \theta=\displaystyle\frac{\pi}{4}\;or\;\displaystyle\frac{-\pi}{4}$
Option $A$ is correct

The position of the point $(1, 3)$ with respect to the ellipse $4x^2+9y^2-16x-54y+61=0$.

  1. Outside the ellipse

  2. On the ellipse

  3. On the major axis

  4. On the minor axis


Correct Option: A

The point at shortest distance from the line x+y=7 and lying on an ellipse $x^2 + 2y^2 =6$, has coordinates

  1. ($\sqrt{2}, \sqrt{2}$)

  2. ($0, \sqrt{3}$)

  3. ($\sqrt{5}, \dfrac{1}{\sqrt{2}}$)

  4. (2, 1)


Correct Option: A

Which of the following points is an exterior point of the ellipse $\displaystyle 16 x^{2} + 9y^{2} - 16x - 32 = 0$.

  1. $\displaystyle \left ( \frac{1}{2}, : 2 \right )$

  2. $\displaystyle \left ( \frac{1}{4}, : 2 \right )$

  3. $\displaystyle \left ( 3, : 2 \right )$

  4. none of these


Correct Option: B,C
Explanation:

Let   $S = \displaystyle 16 x^{2} + 9y^{2} - 16x - 32 $
Now $S(\dfrac12,2)=4+36-8-32 = 0 \Rightarrow $ point on the ellipse.
$S(\dfrac14,2) = 1+36-4-32> 0 \Rightarrow $ point is exterior to the ellipse.
$S(3,2) = 144+36-48-32>0 \Rightarrow $ point is exterior to the ellipse.

An ellipse with foci $(0,\pm 2)$ has length of minor axis as $4$ units. Then the ellipse will pass through the point

  1. $\left( 2,\sqrt { 2 } \right) $

  2. $\left( \sqrt { 2 } ,2 \right) $

  3. $\left( 2,2\sqrt { 2 } \right) $

  4. $\left( 2\sqrt { 2 } ,2 \right) $


Correct Option: B
Explanation:

Let $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1(a<b)\quad $ is the equation of ellipse, foci $(0,\pm 2)$
(be $=2$)
Given: $2a=4\Rightarrow a=2$
${ e }^{ 2 }=1-\cfrac { { a }^{ 2 } }{ { b }^{ 2 } } \Rightarrow { b }^{ 2 }{ e }^{ 2 }={ b }^{ 2 }-{ a }^{ 2 }\quad $
$\quad 4={ b }^{ 2 }-4\Rightarrow { b }^{ 2 }=8$
$\therefore$ equation of ellipse is $\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ 8 } =1\quad $
It passes through $\left( \sqrt { 2 } ,2 \right) $

$C: x^{2}+y^{2}=9$, $\displaystyle E: \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, $L: y=2x$

Let $L$ intersect $x=1$ at point $R$. Then which of the following is correct :
  1. $R$ lies inside both $C$ and $E$

  2. $R$ lies outside both $C$ and $E$

  3. $R$ lies on both $C$ and $E$

  4. $R$ lies inside $C$ but outside $E$


Correct Option: D
Explanation:

$y=2x$, intersects $x=1$ at $(1,2)$
Coordinate of $R$ are $(1,2)$
$C(1,2)=1+22-9<0$ Since $C(1,2)$ is $<0, R $ lies inside $C$
$E(1,2)=\dfrac{1}9+1-1>0$ Since $E(1,2)$ is $>0, R $ lies outside $E$.

Let a curve satisfying the differential equation $y^2dx+\left(x-\dfrac{1}{y}\right)dy=0$ which passes through $(1, 1)$. If the curve also passes through $(k, 2)$, then value of k is?

  1. $\dfrac{1}{2}-\dfrac{1}{\sqrt{e}}$

  2. $\dfrac{3}{2}+\dfrac{1}{\sqrt{e}}$

  3. $\dfrac{3}{2}-\dfrac{1}{\sqrt{e}}$

  4. $\dfrac{1}{2}+\dfrac{1}{\sqrt{e}}$


Correct Option: C
Explanation:

$y^2dx+\left(x-\dfrac{1}{y}\right)dy=0$
$\Rightarrow \dfrac{dx}{dy}+\dfrac{x}{y^2}=\dfrac{1}{y^3}$
Integrating factor (I.F.)$=e^{-\dfrac{1}{y}}$
Now $x.e^{-\dfrac{1}{y}}=\displaystyle\int e^{-\dfrac{1}{y}}\dfrac{1}{y^3}dy$
Put $-\dfrac{1}{y}=y$
$x.e^t=\displaystyle\int e^t(-t)dt$
$\Rightarrow x.e^t=-(t.e^t-e^t)+c$
$\Rightarrow e^{-\dfrac{1}{y}}=e^{-\dfrac{1}{y}}\left(1+\dfrac{1}{y}\right)+c$
$\Rightarrow x=1+\dfrac{1}{y}+c.e^{\dfrac{1}{y}}$
it passes through point $(1, 1)$
$\therefore c=-\dfrac{1}{e}$
Equation of curve is
$x=1+\dfrac{1}{y}-e^{\dfrac{1}{y}-1}$
It passes through $(k, 2)$
$\therefore k=1+\dfrac{1}{2}-e^{-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{1}{\sqrt{e}}$.

Let $E$ be the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 4 } =1$ and $C$ be the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$. Let $P$ and $Q$ be the points $(1,2)$ and $(2,1)$ respectively. Then

  1. $Q$ lies inside $C$ but outside $E$

  2. $Q$ lies outside both $C$ and $E$

  3. $P$ lies inside both $C$ and $E$

  4. $P$ lies inside $C$ but outside $E$


Correct Option: D
Explanation:

Since $\displaystyle \frac { { 1 }^{ 2 } }{ 9 } +\frac { { 2 }^{ 2 } }{ 4 } -=\frac{1}{9}>0$

$\therefore P(1,2)$ lies outside $E$
Since $\displaystyle \frac { { 2 }^{ 2 } }{ 9 } +\frac { { 1 }^{ 2 } }{ 4 } -1<0$
$\therefore Q(2,1)$ lies inside $E$
Since ${ 1 }^{ 2 }+{ 2 }^{ 2 }-9<0$
$\therefore P(1,2)$ lies inside $C$
Since ${ 2 }^{ 2 }+{ 1 }^{ 2 }-9<0$
$\therefore Q(2,1)$ also lies inside $C$
$\therefore P$ lies inside $C$ but outside $E$.

Find the equation of the ellipse whose eccentricity is $\dfrac{4}{5}$ and axes are along the coordinate axes and foci at $(0, \pm 4)$.

  1. $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$

  2. $\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$

  3. $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$

  4. $\dfrac{x^2}{9}+\dfrac{y^2}{36}=1$


Correct Option: A
Explanation:

Let the required equation of the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.


According to the problem, the coordinates of the foci are $(0, \pm 4)$.

We know, coordinates of foci are $(0, \pm be)$.

Therefore, $be =4$

$b\left (\dfrac{4}{5}\right )=4$

$b=5$

$b^2=25$

Now, $a^2=b^2(1-e^2)$

$a^2=5^2\left (1-\dfrac{16}{25}\right )$

$a^2=9$

Thus, the required equation of ellipse is $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$.

The point $(4, -3)$ with respect to the ellipse $4x^2+5y^2=1$.

  1. lies on the curve

  2. lies inside the curve

  3. lies outside the curve

  4. lies focus of the curve


Correct Option: C
Explanation:

$Equation\quad of\quad ellipse:\quad 4{ x }^{ 2 }+5{ y }^{ 2 }=1\ Putting\quad the\quad point\quad (4,-3)\quad on\quad the\quad ellipse\quad we\quad get:\ \quad =4{ (4) }^{ 2 }+5{ (-3) }^{ 2 }-1\ =\quad 64+45-1=28>0\ \therefore \quad The\quad point\quad lies\quad outside\quad the\quad ellipse.$


Option [C]

Consider the ellipse with the equation $x^{2}+3y^{2}-2x-6y-2=0.$ The eccentric angle of a point on the ellipse at a distance 2 units from the contra of the ellipse is

  1. $\dfrac{\pi }{4}$

  2. $\dfrac{\pi }{2}$

  3. $\dfrac{\pi }{6}$

  4. $\dfrac{\pi }{3}$


Correct Option: A
Explanation:

We have,  $x^{2}+3y^{2}-2x-6y-2=0.$

$\Rightarrow$ $x^{2}-2x+1-1+3y^{2}-6y+3-3-2=0.$

$\Rightarrow$ $(x-1)^{2}+3(y-1)^{2}=6$

$\Rightarrow$ $(x-1)^{2}+3(y-1)^{2}=6$ which becomes $x^{2}+3y^{2}=6$ on shifting the origin to $(1, 1)$. Any point with eccentric angle $\theta $ is $(\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$ 

$\Rightarrow$ $4=6cos^{2}\theta +2sin^{2}\theta \Rightarrow 4cos^{2}\theta =2\Rightarrow cos\theta =\pm \dfrac{1}{\sqrt{2}}$

$\Rightarrow$ Hence $\theta =\dfrac{\pi }{4}$ 

Find the set of value(s) of $\alpha$ for which the point $\left ( 7\,-\, \displaystyle \frac{5}{4}\alpha,\,\alpha \right )$ lies inside the ellipse $\displaystyle \frac{x^2}{25}\,+\,\frac{y^2}{16}\,=\, 1.$

  1. $ \displaystyle\left( \frac{17}{5} \dfrac{12}{5}\right) $

  2. $ \left(\dfrac{12}{5},\dfrac{16}{5}\right) $

  3. $ \dfrac{-16}{5} $

  4. None of these


Correct Option: B
Explanation:

$\displaystyle \frac{x^2}{25}\,+\,\frac{y^2}{16}\,=\, 1$


Since point $(7\,-\, \displaystyle \frac{5}{4}\alpha,\,\alpha)$ lies inside the ellipse

$\therefore S _1\,<\,0$

$\Rightarrow 16 (7\,-\, \displaystyle \frac{5}{4}\alpha)^2\,+\, 25.\alpha^2\,<\,400$

$\Rightarrow\,(28\, -\,5\alpha)^2\,+\, 25\alpha^2\,<\, 400$

$\Rightarrow\, 50\alpha^2\, -\, 280\,\alpha\,+\, 384\, < \,0$

$\Rightarrow\, 25\alpha^2\, -\, 140\,\alpha\,+\, 192\, < \,0$

$\Rightarrow (5\alpha-12)(5\alpha-16)<0$

$\Rightarrow \, \alpha \, \in \, \left( \dfrac { 12 }{ 5 } ,\, \dfrac { 16 }{ 5 }  \right) $

$\mathrm{A}$ssertion ($\mathrm{A}$): The point $(5,-2)$ lies outside the ellipse $24x^{2}+7y^{2}=12$.
Reason (R): lf the point $(x _{1},y _{1})$ lie outside the ellipse $\mathrm{S}=0$ then $S _{11}>0$ 

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true but R is not coorect explanation of A

  3. A is true but R is false

  4. A is false but R is true


Correct Option: A
Explanation:

$24x^{2}+7y^{2}-12=5$

$S(5,-2)=24\times 25+7\times 4-12$

$S(5,-2)>0$

$\therefore $ lies outside the ellipse.
$S=0$
$S(x _{1}y _{1})>0$ the point is outside the ellipse.

The point $(2\cos \theta , 3\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$.

  1. outside

  2. inside

  3. on the periphery of

  4. on the auxillary of


Correct Option: C
Explanation:

Given point is $\left ( 2\cos\theta, 3\sin\theta  \right )$


Substituting given point in the ellipse equation:
$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1$
$= \dfrac{4\cos^{2}\theta}{4}+\dfrac{9\sin^{2}\theta }{9}-1$
$= \cos^{2}\theta + \sin^{2}\theta -1$
$= 1-1=0$

$\therefore$ Given point satisfies the ellipse 
$\Rightarrow$ point lie on periphery of ellipse.

The distance of point '$\theta$' on the ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1$ from a focus is:

  1. $a(e + \cos \theta)$

  2. $a(e - \cos \theta)$

  3. $a(1 + e \cos \theta)$

  4. $a(1 + 2e \cos \theta)$


Correct Option: C
Explanation:
Given equation of ellipse is
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
Any point on the ellipse will be $(a cos\theta, b sin \theta)$
$P=(acos\theta, b sin \theta)$
Centre of the ellipse is $(0,0)$
Ellipse is parallel to horizontal axis
Foci of the ellipse is
$F=(h-ae,k)$ if $(h,k) $ is the centre
$F=(0-ae,0)=(-ae,o)$
Distance $FP=\sqrt{(-ae-cos\theta)^2+(bsin\theta-0)^2}$
$=a \sqrt{(e^2+cos^2 \theta+2ecos\theta+(1-e^2)sin \theta-0)^2}$
$a\sqrt{(e^2+cos^2 \theta+2ecos\theta+sin^2 \theta-e^2sin^2\theta)}$
$a\sqrt{(1+2ecos\theta+e^2(1-sin^2\theta)}$
$a\sqrt{(1+2ecos \theta+e^2cos^2\theta)}$
$a\sqrt{(1+ecos\theta)^2}$
$FP=a(1+ecos\theta)$



$(2,3)$ lies _______  the ellipse $16 x^{2} + 9y^{2} - 16x - 32 = 0$

  1. Inside

  2. Outside

  3. On

  4. None of the above


Correct Option: B
Explanation:
Given ellipse $16x^2+9y^2-16x-32=0$
Let $S=16x^2+9y^2-16x-32=0$
Put $(2,3) $ in $S$ we get
$S(2,3)=16(2)^2+9(3)^2-16(2)-32=81 $
$S(2,3)>0$
So it $(2,3) $ lies outside the ellipse.

The point $(4\cos \theta , 4\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$

  1. outside

  2. inside

  3. on the periphery

  4. on the auxiliary circle


Correct Option: D
Explanation:

Equation of ellipse is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 9 } =1$

Equation of auxiliary circle of ellipse is
${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\ { x }^{ 2 }+{ y }^{ 2 }=16$
Substiuting $(4\cos\theta,4\sin\theta$) in the equation, we get
${ (4\cos { \theta  } ) }^{ 2 }+{ (4\sin { \theta  } ) }^{ 2 }=16\ 16(\cos ^{ 2 }{ \theta  } +\sin ^{ 2 }{ \theta  } )=16\ 16=16$
The point satisfies the equation of auxiliary circle.
So, option D is the correct.

$(3,2)$ lies _______  the ellipse $16 x^{2} + 9y^{2} - 16x - 32 = 0$

  1. Inside

  2. Outside

  3. On

  4. None of the above


Correct Option: B
Explanation:

Ellipse : $16{ x }^{ 2 }+9{ y }^{ 2 }-16x-32=0$

Let $S=16{ x }^{ 2 }+9{ y }^{ 2 }-16x-32$
Putting point (3,2) in S we get
         S(3,2) $=16\times { (3) }^{ 2 }+9\times { (2) }^{ 2 }-16\times 3-32$
                   $=144+36-48-32$
                   $=180-80$
                   $=100$
          $S(3,2) > 0$
Hence, point lies outside ellipse.

The point $(1,1)$ lies ____________ the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$

  1. outside

  2. inside

  3. on the periphery

  4. on the auxillary circle


Correct Option: B
Explanation:

We will substitute the given point in ellipse equation

$\Rightarrow$ $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1$
$=\dfrac{1}{4}+\dfrac{1}{9}-1$
$\Rightarrow  \dfrac{13}{36}-1$
$\Rightarrow \dfrac{-23}{36}<0$

By Substituting the point, we are getting less than $0$ 

$\therefore $ point lies inside the ellipse .

The distance of a point $(\sqrt 6 \cos \theta, \sqrt 2 \sin \theta)$ on the ellipse $\dfrac {x^2}{6} + \dfrac {y^2}{2}=1$ from the centre is $2$, if:

  1. $\theta =\dfrac {\pi}{2}$

  2. $\theta =\dfrac {3\pi}{2}$

  3. $\theta =\dfrac {5\pi}{2}$

  4. $\theta =  \dfrac{\pi}{4}$


Correct Option: D
Explanation:

$P(\sqrt { 6 } \cos { \theta  } ,\sqrt { 2 } \sin { \theta  } )$

$Centre(0,0)$
$PC=\sqrt { 6\cos ^{ 2 }{ \theta  } +2\sin ^{ 2 }{ \theta  }  } $
$2=\sqrt { 4\cos ^{ 2 }{ \theta  } +2 } $
$2=4\cos ^{ 2 }{ \theta  } $
$\cos { \theta  } =\pm \dfrac { 1 }{ \sqrt { 2 }  } $
$\theta =\dfrac { \pi  }{ 4 } $

State the following statement is True or False
Let $\dfrac {x^2}{9}+\dfrac{y^2}{16}=1$ then
$(\sqrt 2,\sqrt {24})$ is inside the given ellipse.

  1. True

  2. False


Correct Option: B
Explanation:

Equation of ellipse,  $\dfrac { { x }^{ 2 } }{ 9 } +\dfrac { { y }^{ 2 } }{ 16 } =1$

Putting point $\left( \sqrt { 2 } ,\sqrt { 24 }  \right) $ in equation of ellipse
     $=\dfrac { 2 }{ 9 } +\dfrac { 24 }{ 16 } -1$
     $=\left( \dfrac { 31 }{ 18 } -1 \right) >0$
Hence point lies outside ellipse.
False

Let $\dfrac {(x-3) ^2}9+\dfrac {(y-4) ^2}{16}=1$ then   $(0,0)$ is

  1. On the ellipse.

  2. Outside the ellipse.

  3. Inside the ellipse.

  4. None of the above.


Correct Option: B
Explanation:

Equation of ellipse : $\dfrac { { \left( x-3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( y-4 \right)  }^{ 2 } }{ 16 } -1=0$

putting point (0,0) in above ellipse,
          $=\dfrac { { \left( -3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( -4 \right)  }^{ 2 } }{ 16 } -1$
          $=1+1-1$
          $=1>0$
Hence, point lies outside ellipse.

Let $5x^2+7y^2=140$, then $(3,-4)$ is:

  1. Outside the ellipse

  2. Inside the ellipse

  3. On the ellipse

  4. Data insufficient


Correct Option: A
Explanation:

Equation of ellipse :   ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

putting point $(3,-4)$ in equation of ellipse
     $=5\times { 3 }^{ 2 }+7{ \left( -4 \right)  }^{ 2 }-140$
     $=45+112-140$
     $=17(>0)$
Hence, $(3,-4)$ is outside the ellipse.

Let $5x^2+7y^2=140$, then Position of $(4,-3)$ relative to the ellipse is

  1. Inside the ellipse.

  2. Outside the ellipse.

  3. On the ellipse.

  4. None of the above.


Correct Option: B
Explanation:

Equation of ellipse : ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

putting point (4,3) in equation
     $=5\times { \left( 4 \right)  }^{ 2 }+7{ \left( -3 \right)  }^{ 2 }-140$
     $=80+63-140$
     $=3\left( >0 \right) $
Hence, point is outside the ellipse.

Let $\dfrac {(x-3) ^2}9+\dfrac {(y-4) ^2}{16}=1$ then   $(3,4)$ is 

  1. Inside the ellipse

  2. Outside the ellipse

  3. On the ellipse

  4. Centre of the ellipse


Correct Option: A,D
Explanation:

If equation of ellipse is $\dfrac { { \left( x-h \right)  }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { \left( y-k \right)  }^{ 2 } }{ { b }^{ 2 } } =1$,  centre $\left( h,k \right) $

for ellipse   $\dfrac { { \left( x-3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( y-4 \right)  }^{ 2 } }{ 16 } =1$
            Centre of ellipse (3,4)
and centre is inside ellipse only.

Let $5x^2+7y^2=140$, then $(0,0)$ is: 

  1. Inside the ellipse

  2. On the ellipse

  3. Outside the ellipse

  4. Centre of the ellipse


Correct Option: A,D
Explanation:

Equation of ellipse    ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

   Dividing both sides by $140$
           $\dfrac { { x }^{ 2 } }{ 28 } +\dfrac { { y }^{ 2 } }{ 20 } =1$
Comparing with $\dfrac { { \left( x-h \right)  }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { \left( y-k \right)  }^{ 2 } }{ { b }^{ 2 } } =1$
          $(h,k)\ =\ (0,0)$
Hence (0,0) is centre of given ellipse and it inside the ellipse.

Let $5x^2+7y^2=140$, then $(\sqrt {14},\sqrt {10})$ is:

  1. Outside the ellipse

  2. Inside the ellipse

  3. On the ellipse

  4. Centre of the ellipse


Correct Option: C
Explanation:

          ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$          (equation of ellipse)

putting $\left( \sqrt { 14 } ,\sqrt { 10 }  \right) $ in equation
            $=5\times { \left( \sqrt { 14 }  \right)  }^{ 2 }+7\times { \left( \sqrt { 10 }  \right)  }^{ 2 }-140$
            $=70+70-140$
            $=0$
Hence, point lie on the ellipse.

If $P=(x, y), F _1=(3, 0), F _2=(-3, 0)$ and $  16x^2+25y^2=400$, then $PF _1+PF _2$ equals

  1. $8$

  2. $6$

  3. $10$

  4. $12$


Correct Option: C
Explanation:

Given equation is $16x^2+25y^2=400$
The ellipse can be written as, $\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1$
Here $a^2=25, b^2=16$,  but  $b^2=a^2(1-e^2)$
$\Rightarrow \dfrac {16}{25}=1-e^2$
$\Rightarrow e^2=1-\dfrac {16}{25}=\dfrac {9}{25} $
$ \Rightarrow e=\pm \dfrac {3}{5}$
Foci of the ellipse are $(+ae, 0)$ $=$ $ (\pm 3, 0)$, i.e., $F _1$ and $F _2$.
We have $PF _1+PF _2=2a=10$ for every point $P$ on the ellipse.

The position of the point $(1, 2)$ relative to the ellipse $2x^{2} + 7y^{2} = 20$ is

  1. outside the ellipse

  2. inside the ellipse but not at the focus

  3. on the ellipse

  4. at the focus


Correct Option: A
Explanation:

$f\left( x,y \right) ={ 2x }^{ 2 }+7{ y }^{ 2 }-20$

Point will be outside of ellipse if $f\left( 1,2 \right) >0$

on ellipse if $f\left( 1,2 \right) =0$

will be inside if $f\left( 1,2 \right) <0$

$ f\left( 1,2 \right) ={ 2\times 1 }^{ 2 }+7{ \times 2 }^{ 2 }-20=10$

Here we see that $f\left( 1,2 \right) >0$ so point will be outside of ellipse

So correct answer will be option A

The minimum distance of origin from the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is $(a>0,b>0)$

  1. a-b

  2. a+b

  3. 2a+2b

  4. 2(a-b)


Correct Option: B
Explanation:
Let $(a sec\theta,b cosec \theta)$ be a point on the curve, then its diastance  from the origin
$=\sqrt{a^2 sec^2 \theta + b^2 cosec ^2 \theta}$
$\therefore f(\theta)=a^2 sec^2 \theta +b^2 cosec ^2 \theta$
$=a^2+a^2 tan ^2 \theta + b^2 + b^2 cot ^ 2 \theta$
$=a^2+b^2 + a^2 tan ^2 \theta + b^2 cot ^2 \theta$
$\geq a^2+b^2+2 \sqrt{a^2 b^2}=(a+b)^2$
$\therefore$ minimum value of $f(\theta)=(a+b)^2$
$\therefore$ minimum value of $\sqrt{a^2 sec^2 \theta + b^2 cosec^2 \theta }$ is $a+b$

If $a$ and $c$ positive real number and the ellipse $\dfrac { { x }^{ 2 } }{ { 4c }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { c }^{ 2 } } =1$ has four distinet points in common with the circle ${ x }^{ 2 }+{ y }^{ 2 }=9{ a }^{ 2 }$, then

  1. $6ac+9{ a }^{ 2 }-2{ c }^{ 2 }>0$

  2. $6ac+9{ a }^{ 2 }-2{ c }^{ 2 }<0$

  3. $9ac-9{ a }^{ 2 }-2{ c }^{ 2 }<0$

  4. $9ac-9{ a }^{ 2 }-2{ c }^{ 2 }>0$


Correct Option: C

An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is $2/3$ then the eccentricity of the ellipse is:

  1. $\dfrac{2\sqrt{2}}{3}$

  2. $\dfrac{\sqrt{5}}{3}$

  3. $\dfrac{8}{9}$

  4. $\dfrac{2}{3}$


Correct Option: A

The segment of the tangent at the point P to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, intercepted by the auxiliary circle subtends a right angle at the origin. If the eccentricity of the ellipse is smallest possible, then the point P can be

  1. $(0,ae)$

  2. $(a,0)$

  3. $(-a,0)$

  4. $(0,-b)$


Correct Option: D
Explanation:

Equation of tangent at $P (a cos \theta, b sin \theta)$ is
$\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$
$\therefore$ Equation of the lines OA and OB is
$x^{2}+y^{2}-a^{2}\left(\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}\right)^{2}=0$
since the lines OA and OB are perpendicular
$\therefore$ $1- \cos ^{2}\theta + 1 - \frac{a^{2}}{b^{2}} \sin^{2}\theta=0$
i.e $\sin^{2}\theta + 1-\frac{a^{2}}{b^{2}} \sin^{2}\theta$
i.e $\frac{b^{2}}{a^{2}}=\frac{sin^{2}\theta}{1+ \sin^{2} \theta}$
$\therefore e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{\sin^{2} \theta}{1+\sin^{2}\theta}=\frac{1}{1+\sin^{2}\theta}$
$\therefore$ e is least if $\theta = \pm \frac{\pi}{2}$
$\therefore$ the point P is $(0,\pm b)$

If one end of the diameter of the ellipse $4x^2+y^2=16$ is $(\sqrt 3, 2)$, then the other end is:

  1. $(\sqrt 3, 2)$

  2. $(-\sqrt 3, 2)$

  3. $(-\sqrt 3, -2)$

  4. $(\sqrt 3, -2)$


Correct Option: C
Explanation:

The center of ellipse is $(0,0)$

The diameter of ellipse will pass through the center of ellipse and center will divide the diameter into two halfs
One end of diameter is $(\sqrt3,2) $ , Let the other end be $(h,k)$
We have $h+\sqrt3=0$ and $k+2=0$
$\Rightarrow h=-\sqrt3,k=-2$
Therefore option $C$ is correct

The point P on the ellipse $4x^2+9y^2=36$ is such that the area of the $\Delta PF _1F _2=\sqrt{10} Sq$ units, where $F _1.F _2$ are Foci. Then P has the coordinates

  1. $(\pm\dfrac{3}{\sqrt{2}},\sqrt{2})$

  2. $(\dfrac{3}{2},2)$

  3. $(\dfrac{-3}{2},-2)$

  4. NONE


Correct Option: A
Explanation:

$4x^2+9y^2=36\Rightarrow \dfrac {x^2}9+\dfrac {y^2}4=1$

So, $a=3,b=2$

In an ellipse, Focal points, $F=(\pm\sqrt {a^2-b^2},0)=(\pm\sqrt 5,0)$
So, $F _1F _2=2\sqrt 5$ which is the base of the triangle $PF _1F _2$

An arbitrry point P on ellipse is $(acos\theta,b\sin\theta)$
So, $Height\ of\ \triangle PF _1F _2=b\sin\theta=2\sin\theta$

ie, $Area=\dfrac 12base\times height=\dfrac 12\times 2\sqrt 5\times2\sin\theta=2\sqrt 5\sin\theta=\sqrt {10}$
$\Rightarrow \sin\theta=\dfrac 1{\sqrt 2}$
$\Rightarrow \cos\theta=\pm\dfrac 1{\sqrt 2}$
And $P=(a\cos\theta,b\sin\theta)=(\pm\dfrac 3{\sqrt 2},\sqrt 2)$

So, Option$ A$ has one of the anwers.

Which of the following is an (x,y) coordinate pair located on the ellipse $4x^2 + 9y^2 = 100$?

  1. $(1, 3.5)$

  2. $(1.4, 3.2)$

  3. $(1.9, 2.9)$

  4. $(2.3, 3.1)$

  5. $(2.7, 2.6)$


Correct Option: B
Explanation:
  • Substitute each and every point on given ellipse and see which will satisfy the equation
  • Take point $(1.4,3.2)$ , when we substitute we get $4(1.96)+9(10.24) = 100$
  • Therefore correct answer is option $B$
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