Tag: position of point wrt ellipse

Questions Related to position of point wrt ellipse

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 

position of point wrt ellipse ellipse maths

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

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

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

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

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

position of point wrt ellipse ellipse maths

  1. inside

  2. outside

  3. on

  4. None of the above

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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

position of point wrt ellipse ellipse maths

  1. $2\pi/3$

  2. $\pi/3$

  3. $4\pi/3$

  4. $none\ of\ these$

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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:

position of point wrt ellipse ellipse maths

  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

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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 

position of point wrt ellipse ellipse maths

  1. $5$

  2. $7$

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

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

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Correct Option: C

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

position of point wrt ellipse ellipse maths

  1. inside

  2. outside

  3. on

  4. None of the above

Show answer
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.

position of point wrt ellipse ellipse maths

  1. True

  2. False

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Correct Option: B

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

position of point wrt ellipse ellipse maths

  1. Outside

  2. Inside

  3. On the ellipse

  4. None of the above

Show answer
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 

position of point wrt ellipse ellipse maths

  1. outside

  2. inside

  3. on

  4. None of the above

Show answer
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 $\dfrac{x^2}{4}+\dfrac {y^2}{3}=1$

position of point wrt ellipse ellipse maths

  1. Inside 

  2. Outside

  3. On the Ellipse 

  4. None.

Show answer
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