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Tracing of the parabola - class-XII

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The solution of \$frac{{dy}}{{dx}} = \frac{{ax + h}}{{by + k}}$ represents a parabola

  1. a=0,b=0

  2. a=1, b=2

  3. a=0, b=0

  4. a=2, b=1


Correct Option: A

The equation $y^2+3 =2( 2x +y)$ represents a parabola with vertex at 

  1. $\left(\dfrac{1}{2}, 1\right) $ and axis parallel to $y$-axis

  2. $\left(\dfrac{1}{2}, 1\right) $ and axis parallel to $ x$-axis

  3. $\left(\dfrac{1}{2}, 1\right) $ and focus at $\left(\dfrac{3}{2}, 1\right)$

  4. $\left(1, \dfrac{1}{2},\right) $ and focus at $\left(\dfrac{3}{2}, 1\right)$


Correct Option: B,C
Explanation:

$y^2+3=2(2x+y)$ represents parabola.


$y^2+3=4x+2y$


$y^2-2y+3=4x$

$y^2-2y+1+3=4x+1$

$(y-1)^2=4x-2$

$(y-1)^2=4(x-\dfrac{1}{2})$

So, the vertex of parabola$=\left(\dfrac{1}{2},1\right)$ and axis is parallel to x axis.

$a=1$

Focus$=\left(\dfrac{1}{2}+1,1\right)$

               $=\left(\dfrac{3}{2},1\right)$

If the equation of parabola is ${x}^{2}=-9y$, then the equation of the directrix and the length of latus rectum are 

  1. $y=-\dfrac {9}{4}, 8$

  2. $x=\dfrac {-9}{4}, 9$

  3. $y=\dfrac {9}{4}, 9$

  4. $None\ of\ these$


Correct Option: C
Explanation:
The given equation is of the form ${x}^{2} = - 4ay$, where a is positive.
Therefore, the focus is on the y-axis in the negative direction and parabola opens downwards.
Therefore,
Given equation of parabola-
${x}^{2} = -9 y ..... \left( 1 \right)$
Standard equation of parabola-
${x}^{2} = -4ay ..... \left( 2 \right)$
Comparing ${eq}^{n} \left( 1 \right) \& \left( 2 \right)$, we have
$a = \cfrac{9}{4}$
As we know that, for parabola in the form ${x}^{2} = -4ay$, equation of directrix for parabola is-
$y = a$
$\therefore$ For ${x}^{2} = -9y$, equation of directrix is-
$y = \cfrac{9}{4}$
Length of latus rectum, $l = 4a$
As $a = \cfrac{9}{4}$,
$\therefore \; l = 4 \times \cfrac{9}{4} = 9$
Hence, the equation of directrix will be $y = \cfrac{9}{4}$ and the lngth of latus rectum will be 9.

If $\displaystyle \left ( 2,0 \right )$ is the vertex and $y -$ axis the directrix of a parabola,find the coordinates of focus. 

  1. Focus is $\displaystyle \left ( 2,0 \right )$

  2. Focus is $\displaystyle \left ( 4,0 \right )$

  3. Focus is $\displaystyle \left ( 8,0 \right )$

  4. Focus is $\displaystyle \left ( -4,0 \right )$


Correct Option: B
Explanation:

Distance of vertex from directrix $y -$ axis is $2$ and we know it
is half the distance of focus from directrix.
$\displaystyle \therefore $ Focus is $\displaystyle \left ( 4,0 \right ).$

The focal distance of a point $P$ on the parabola $y^2=12x$ if the ordinate of $P$ is $6$, is

  1. $12$

  2. $6$

  3. $3$

  4. $9$


Correct Option: B
Explanation:

Given parabola is $y^2=12x$    ....$(i)$
Here $a=3$
For point $P(x, y), y=6$
This point lie on the parabola 
$\therefore (6)^2=12x\Rightarrow x=3$
Now, focal distance of point $P$ is $x+a=6$

The equation of the conic with focus $\displaystyle S \left( \frac{3}{2}, 0 \right) $ and the directrix 2x + 3 = 0 having eccentricity 1, is

  1. $y^2 = 4x$

  2. $y^2 = 5x$

  3. $y^2 = 6x$

  4. $y^2 = 8x$


Correct Option: C

The locus of the points which are equidistant from $(-a, 0)$ and $x=a$ is

  1. $y^2=4ax$

  2. $y^2+4ax=0$

  3. $x^2+4ay=0$

  4. $x^2-4ay=0$


Correct Option: B
Explanation:
Let $(h,k)$ be any point on the curve.
Distance of this point from $(-a,0)=\sqrt{(h-(-a))^2+(k-0)^2}$
Distance of point $(h,k)$ from the line $x-a=0$ is $\dfrac{h-a}{1}=h-a$
Point $(h,k)$ is equidistant from $(-a,0)$ and the line $x-a=0$
$\implies  \sqrt{(h-(-a))^2+(k-0)^2}=h-a$

$\implies \sqrt{(h+a)^2+(k-0)^2}=h-a$

Squaring the above equation, we get
$\implies (h+a)^2+(k-0)^2=(h-a)^2$
$\implies h^2+a^2+2ah+k^2=h^2+a^2-2ah$

$\implies k^2=-4ah$
Subtitute $k=y$ and $h=x$, we get

$\implies y^2=-4ax$
$\implies y^2+4ax=0$
So, the answer is option (B)


Find the equation of the parabola whose focus is $S(3,5)$ and vertex is $A(1,3)$.

  1. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 3} \right)}^2} + {{\left( {y - 5} \right)}^2}} \right]\end{array}$

  2. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 6} \right)}^2} + {{\left( {y - 6} \right)}^2}} \right]\end{array}$

  3. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 11} \right)}^2} + {{\left( {y - 11} \right)}^2}} \right]\end{array}$

  4. $\begin{array}{}\ \Rightarrow \left|  \right| =  {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 7} \right)}^2} + {{\left( {y - 7} \right)}^2}} \right]\end{array}$


Correct Option: A
Explanation:

Slope of axis$=\frac{{5 - 3}}{{3 - 1}} = \frac{2}{2} = 1$
Slope of directrix$=-1$
equation of tangent at vertex $A$
$\begin{array}{l}Pt(1,3),m =  - 1\ \Rightarrow y - 3 =  - 1\left( {x - 1} \right)\ \Rightarrow x + y - 4 = 0\end{array}$
equation of directrix
$\begin{array}{l}x + y = \lambda \a = SA\ = \sqrt {4 + 4}  = 2\sqrt 2 \end{array}$
$A$ is midpoint of $PS$
$\begin{array}{l}\frac{{n + 3}}{2} = 1,\frac{{k + 5}}{2} = 3\ \Rightarrow n =  - 1,k = 1\end{array}$
$(-1,1)$ lies on directrix
$-1+1=\lambda=0$
equation of diectrix: $L:y+x=0$
$\begin{array}{l}QO = QS\ \Rightarrow \left| {\frac{{l + m}}{{\sqrt 2 }}} \right| = \sqrt {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \ \Rightarrow {\left( {l + m} \right)^2} = 2\left[ {{{\left( {l - 3} \right)}^2} + {{\left( {m - 5} \right)}^2}} \right]\ \Rightarrow {\left( {x + y} \right)^2} = 2\left[ {{{\left( {x - 3} \right)}^2} + {{\left( {y - 5} \right)}^2}} \right]\end{array}$

eccentricity of the conic $25x^2-9y^2 = 225$, are

  1. $\dfrac25$

  2. $\dfrac45$

  3. $\dfrac13$

  4. $\dfrac15$


Correct Option: B

The equation $(13x - 1)^{2} + (13y - 1)^{2} = k(5x - 12y + 1)^{2}$ will represent a parabola if

  1. $k = 2$

  2. $k = 81$

  3. $k = 169$

  4. $k = 1$


Correct Option: D
Explanation:
We know that the general equation of the ellipse is 
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}=e\dfrac{{\left(lh+mk+n\right)}^{2}}{{l}^{2}+{m}^{2}}$        ................$(1)$

${\left(13x-1\right)}^{2}+{\left(13y-1\right)}^{2}=k{\left(5x-12y+1\right)}^{2}$

$\Rightarrow\,{13}^{2}{\left(x-\dfrac{1}{13}\right)}^{2}+{13}^{2}{\left(y-\dfrac{1}{13}\right)}^{2}=k{\left(5x-12y+1\right)}^{2}$

$\Rightarrow\,169\left[{\left(x-\dfrac{1}{13}\right)}^{2}+{\left(y-\dfrac{1}{13}\right)}^{2}\right]=k{\left(5x-12y+1\right)}^{2}$

$\Rightarrow\,{\left(x-\dfrac{1}{13}\right)}^{2}+{\left(y-\dfrac{1}{13}\right)}^{2}=\dfrac{k}{169}{\left(5x-12y+1\right)}^{2}$       ...........$(2)$


A parabola has its eccentricity $e=1$

Comparing equations $(1)$ and $(2)$ we get

$\Rightarrow\,k=1$

$\therefore\,k=1$

If the eqn of directrix to the parabola $x^{2}+4y-6x+\lambda=0$ is $y+1=0$, then 

  1. $\lambda=9$

  2. $\lambda=-17$

  3. Focus $(3, -3)$

  4. vertex (3, -2)$


Correct Option: A

The focus of the parabola $y ^ { 2 } = 4 y - 4 x$ is

  1. $( 0,2 )$

  2. $( 1,2 )$

  3. $( 4,2 )$

  4. $( 1,3 )$


Correct Option: A
Explanation:

$\Rightarrow y^2 = 4y - 4x$

$\Rightarrow y^2 - 4y = -4x$
$\Rightarrow y^2 - 4y + 4 = -4x + y$
$\Rightarrow (y - 2)^2 = -4 (x - 1)$
vertex = $(1, 2)$
$\dfrac{1}{4 p} = \dfrac{-1}{4} \rightarrow p = -1$
$\therefore$ Focus = $(-1 + 1, 2 + 0) = (0, 2)$

Equation of the directrix of the parabola whose focus is $(0,0)$ and the tangent at the vertex is $x-y+1=0$ is 

  1. $x-y=0$

  2. $x-y-1=0$

  3. $x-y+2=0$

  4. $x+y-1=0$


Correct Option: A
Explanation:

Putting $x=1,y=2$

$1-2+1=0\-1+1=0$
So, equation of directrix of parabola $x-y=0$

The equation of the directrix of the parabola, $y ^ { 2 } + 4 y + 4 x + 2 = 0$ is -

  1. $x = - 1$

  2. $x = 1$

  3. $x = - \dfrac {3 }{ 2}$

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


Correct Option: D
Explanation:
Given equation of parabola
$y^2+4y+4x+2=0$
or $(y+2)^2+4x+2=0$
or $(y+2)^2=-4(x-12)$
Its directrix  is x12x−12=1=1
or x=32
Hence D is the correct answer.

y2+4y+4x+2=

The equation of directrix of the parabola $(y-2)^{2}=4(x-4)$, is

  1. $x+1=0$

  2. $x=1$

  3. $x=2$

  4. $x=4$


Correct Option: A

The vertex of the parabola $ {4y}^{2} + 12x-12y+39= 0$ is:

  1. $ \left( 5/2,3/2 \right)$

  2. $ \left ( -5/2,3/2 \right) $

  3. $ \left ( -5/2,-3/2 \right )$

  4. $ \left ( 5/2,-3/2 \right )$


Correct Option: A

The focus of the parabola $(y-2)^{2}=20(x+3)$ is:

  1. $(2, 2)$

  2. $(0, 2)$

  3. $(3, -2)$

  4. $(-2, 0)$


Correct Option: A

A parabola is written as $x^{2}=4ay$, its focus and equation of the directrix is:

  1. $(a, 0)$ and $x+a=0$

  2. $(0, a)$ and $y+a=0$

  3. $(0, a)$ and $x-a=0$

  4. $(0, a)$ and $y-a=0$


Correct Option: A

Focus of the parabola $4x^{2}-12x+8y+13=0$ is

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

  2. $\left(\dfrac {3}{2},\ -5 \right)$

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

  4. $\left(\dfrac {3}{2},\ -1 \right)$


Correct Option: A

The locus of the foot of the perpendicular from the focus upon a tangent to the parabola $y^{2}=4ax$ is 

  1. $the\ directrix$

  2. $tangent\ at\ the\ vertex$

  3. $x=a$

  4. $none\ of\ these$


Correct Option: A

Maximum radius of the circle inscribed in parabola ${y}^{2}=4x$ with centre its focus is ..

  1. $8$

  2. $4$

  3. $2$

  4. $5$


Correct Option: A

The locus of the midpoint of the line segment joining the focus to a  moving point on the parabola y$^2$ - 4ax is another parabola with directrix

  1. x = - a

  2. x = a

  3. x = 0

  4. x = $\dfrac{a}{2}$


Correct Option: A

Equation of the directrix of the parabola $4y^2-6x-4y-5=0$ is 

  1. $8x+11=0$

  2. $8x-11=0$

  3. $11x+8=0$

  4. $11x-8=0$


Correct Option: A

The equation of the directrix of the parabola $y= x^2-2x+3$ is 

  1. $y= 1/4$

  2. $y= 1/4$

  3. $y= 7/4$

  4. $y= - 7/4$


Correct Option: B

The focus of the parabola $x^2 -4x+2y+8=0$ is 

  1. $(3/2, -2)$

  2. $(5/2, -2)$

  3. $(2, -3/2, )$

  4. $(2, -5/2, )$


Correct Option: A

The equation of the directrix of the parabolas $x=-2at,\ y=-at^{2},\ t\ \epsilon \ R$ is

  1. $x-a=0$

  2. $y-a=0$

  3. $x+a=0$

  4. $y+a=0$


Correct Option: A

The angle of intersection at the origin to the curves ${ y }^{ 2 }=4x$ and ${ x }^{ 2 }=4y$ is :

  1. $\pi $

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

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

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


Correct Option: D
Explanation:
Given  $y^2=4x$

Differentiating w.r.t. $x$, we get,

$2y\dfrac{dy}{dx}=4$

$\dfrac{dy}{dx}=\dfrac2y.......1$

$x^2=4y$

Differentiating w.r.t. $x$, we get,

$2x=4\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{x}{2}...........2$

So the slope of tangent at $(0,0)$ of $(1)$ is Parallel to $y$ axis 

And the slope of tangent at $(0,0)$ of $(2)$ is Parallel to $x$ axis 

So the angle between them is $\dfrac{\pi}{2}$ 

if the vertex and the focus of the parabola are $(-1, -1) & (2, 3)$ respectively, then the equation of the directrix is

  1. $3x + 2y + 14 = 0$

  2. $3x + 2y - 25 = 0$

  3. $2x - 3y + 10 = 0$

  4. $x - y + 5 = 0$


Correct Option: A

The parametric equation of a parabola is $x=t^{2}+1, y=2t+1$. The Cartesian equation of its directrix is 

  1. $x=0$

  2. $x+1=0$

  3. $y=0$

  4. $none\ of\ these$


Correct Option: A

$TP$ and $TQ$ are tangents to parabola $y^{2}=4x$ and normal at $P$ and $Q$ intersect at a point $R$ on the curve. The locus of the center of the circle circumscribing $\Delta TPQ$ is parabola whose

  1. Vertex is $\left(1,0\right)$.

  2. Foot of directrix is $\left(\dfrac{7}{8},0\right)$

  3. Length of latus-rectum is $\dfrac{1}{4}$.

  4. Focus is $\left(\dfrac{9}{8},0\right)$


Correct Option: A

For parabola $x^{ 2 } + y^{ 2 } + 2xy 6x 2y + 3 = 0$ the focus is

  1. $\left( 1, -1\right)$

  2. $\left( -1, 1\right)$

  3. $\left( 3, 1\right)$

  4. $None of these$


Correct Option: A
Consider the conic $ x^{2}+4y-6x+k=0 $ & $\displaystyle L\Rightarrow y+1=0$ be its directrix
On the basis of above information answer the following question:

The focus of the parabola is

  1. $(-3, 3)$

  2. $(-3, -3)$

  3. $(3, -3)$

  4. None of these


Correct Option: C
Explanation:
$x^2+4y-6x+k=0$
$\implies (x-3)^2=-4\left ( y-(\dfrac{k-9}{4}) \right )$
So, length of latus rectum $=4a=4$
$\therefore a=1$
Distance between vertex and directrix=a
$\therefore\left ( -1-(\dfrac{k-9}{4}) \right )=1$
$\therefore k=1$
So, co-ordinates of vertex are $\left(3,\dfrac{k-9}{4}\right)=(3,-2)$
Therefore, co-ordinates of focus are $(3,-2-a)=(3,-3)$
So, answer is option (C).
If the focus is $ \displaystyle (\alpha, \beta) $ & the directrix is $ \displaystyle ax+by+c=0 $ then the equation of conic whose eccentricity $=e $ is given by $ \displaystyle \left ( x-\alpha \right )^{2}+\left ( y-\beta \right )^{2}=e^{2}\frac{\left ( ax+by+c \right )^{2}}{a^{2}+b^{2}}$. If $e=1$ then conic is called parabola, for $ e < 1 $ (conic is an ellipse) and for $e > 1,$ conic is a hyperbola. 
Now consider the conic
$\displaystyle 169 \left \{\left (x-1  \right )^{2}+\left (y-3  \right )^{2}  \right \}=\left (5x-12y+17  \right )^{2} $ ......$(*)$
On the basis of above information answer the following question:
 The focus of the conic $(*)$ is
  1. $(-1, -3)$

  2. $(1,3)$

  3. $(5, -12)$

  4. $(-1, 3)$


Correct Option: B
Explanation:

A conic is the locus of a point '$P$' which moves in such a way that its distances from a fixed point'$ S $' always bears a constant ratio to its distances from a fixed straight line. 
The fixed point '$S$' is called focus. 

The fixed straight line is called the directrix & the constant ratio is known as eccentricity denoted by $e$.
$\displaystyle \therefore e=\cfrac {PS}{PM}$
Now $\displaystyle 169\left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\left ( 5x-12y+17 \right )^{2}$
$\displaystyle \Rightarrow \left { \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2} \right }=\frac{\left ( 5x-12y+17 \right )^{2}}{5^{2}+(-12)^{2}}=\left ( \frac{5x-12y+17}{13} \right )^{2}$
$\displaystyle \therefore \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=e^{2}\left ( \frac{5x-12+17}{13} \right )$ where $e=1$

Focus of conic $(* )$ is $(1, 3)$

The vertex of the parabola $2((x-1)^2 + (y-2)^2) = (x + y + 3)^2$ is

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

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

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

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


Correct Option: B
Explanation:

Given parabola may be written as,$\displaystyle \sqrt {(x-1)^2+(y-2)^2}=\frac {|x+y+3|}{\sqrt 2}$
$\Rightarrow$  focus is $S=(1,2)$, diretrix is, $x + y + 3 = 0 ....(1)$
We know axis of the parabola passes through focus and perpendicular to the directrix.
Thus equation of axis is,$x -y + 1 = 0 .....(2)$
solving (1) and (2) we get the foot of directrix $P(-2,-1)$
So the vertex of the parabola will be mid point of PS
$\Rightarrow V = \displaystyle \left (-\frac {1}{2}, \frac {1}{2}\right )$

If a point $\mathrm{P}$ moves such that the distance from the point $\mathrm{A} (1, 1)$ and the line $x+y+2=0$ are equal then the locus of $\mathrm{P}$ is equal to

  1. a straightline

  2. a parabola

  3. a pair of st. lines

  4. an ellipse


Correct Option: B
Explanation:
Let the coordinate at $P$ is $(h, k)$

$\therefore \ $ Distance com point $A(1, 1)$ is $AP=\sqrt {(n-1)^2+(k-1)^2}$

$\therefore \ $ Distance com line $x+y+2=0$ is $=\dfrac {h+k+2}{\sqrt {1^2 +1^2}}$

According to equation

$\sqrt {(h-1)^2 +(k-1)^2}=\dfrac {h+k+2}{\sqrt 2}$

Squaring both are

$2\left\{(h-1)^2 +(k-1)^2\right\}=(h+k+2)^2$

$\Rightarrow \ 2(h^2-2h+1+k^2-2k+1)=h^2+k^2+4+2hk+4h+4k$

$\Rightarrow \ 2h^2-4h+2k^2-4k+4=h^2+k^2+4+2hk+4h+4k$

$\Rightarrow \ h^2-hk+k^2=8h+4k$

$\Rightarrow \ (h-k)^2=8(h+k)$

Locus of $P$ is, $(x-y)^2=8(x+y)$ which is a parabola

Option $\to (B)$

If the vertex of the conic $y^{2} - 4y = 4x - 4a$ always lies between the straight lines $x + y = 3$ and $2x + 2y - 1 = 0$ then

  1. $2 < a < 4$

  2. $-\dfrac {1}{2} < a < 2$

  3. $0 < a < 2$

  4. $-\dfrac {1}{2} < a < \dfrac {3}{2}$


Correct Option: B
Explanation:

Vertex of $y^{2} - 4y = 4x - 4a$ is $(a - 1, 2)$
So, $(a - 1 + 2 - 3)(2a - 2 + 4 - 1) < 0$
$(a - 2)(2a + 1) < 0$
$-\dfrac {1}{2} < a < 2$

For the parabola $9x^{2} - 24xy + 16y^{2} - 20x - 15y - 60 = 0$ which of the following is/ are true.

  1. $focus = \left (-\dfrac {43}{25}, -\dfrac {129}{100}\right )$

  2. $focus = \left (\dfrac {43}{25}, \dfrac {129}{100}\right )$

  3. $directrix : 4x + 3y + \dfrac {53}{4} = 0$

  4. $directrix : 4x + 3y - \dfrac {53}{4} = 0$


Correct Option: B,C

Two manually perpendicular tangent of the parabola ${ y }^{ 2 }=4ax$ meet the axis in ${P} _{1}$ and ${P} _{2}$. If $S$ is the focus of the parabola, then $\dfrac { 1 }{ \left( S{ P } _{ 1 } \right)  } +\dfrac { 1 }{ \left( S{ P } _{ 2 } \right)  } $ is equal to :-

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

  2. $\dfrac { 2 }{ a } $

  3. $\dfrac { 1 }{ a } $

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


Correct Option: C
Explanation:

${ y }^{ 2 }=4ax\quad ............(1)$

Let the two mutually perpendicular tangents have slopes ${ m } _{ 1 },{ m } _{ 1 }$ where ${ m } _{ 1 }=\cfrac { 1 }{ { m } _{ 2 } } $
And hence their questions be,
${ T } _{ 1 }:y=mx+\cfrac { a }{ m } ...........(2)$
${ T } _{ 2 }:y=\cfrac { 1 }{ m } x-am...........(3)$
Let ${ P } _{ 1 }\quad $and$\quad { P } _{ 2 }\quad $be$\quad ({ x } _{ 1 },0)&amp; ({ x } _{ 2 },0)$ these points must lie on ${ T } _{ 1 }$and${ T } _{ 2 }$
For ${ T } _{ 1 },\quad o=m({ x } _{ 1 })+\cfrac { a }{ m } $
       $=>{ x } _{ 1 }=-\cfrac { a }{ { m }^{ 2 } } $
For ${ T } _{ 2 },\quad o=-\cfrac { { x } _{ 2 } }{ m } -am$
       $=>{ x } _{ 2 }=-a{ m }^{ 2 }$
So, ${ P } _{ 1 }(\cfrac { -a }{ { m }^{ 2 } } ,0)\quad $and$\quad { P } _{ 2 }(-a{ m }^{ 2 },0)$
Focus $S=(a,o)$
Now $S{ P } _{ 1 }=\sqrt { (a+{ \cfrac { a }{ { m }^{ 2 } }  })^{ 2 }+0 } =(a+\cfrac { a }{ { m }^{ 2 } } )$
         $S{ P } _{ 2 }=\sqrt { (a+{ a{ m }^{ 2 } })^{ 2 }+0 } =(a+a{ m }^{ 2 })$
Now, $\cfrac { 1 }{ { SP } _{ 1 } } +\cfrac { 1 }{ { { SP } _{ 2 } } } =\cfrac { { m }^{ 2 } }{ a{ m }^{ 2 }+a } +\cfrac { 1 }{ a+{ am }^{ 2 } } $
                               $=\cfrac { { m }^{ 2 }+1 }{ a({ m }^{ 2 }+1) } $
                               $=\cfrac { 1 }{ a } $

A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane.

  1. True

  2. False


Correct Option: A
Explanation:

Parabola is a set of all points in a plane that are equidistant from a fixed line (called directrix) and form a fixed point called focus.

The equation of a parabola in its standard form is $x^3=4ay.$

  1. True

  2. False


Correct Option: B
Explanation:

The given statement is wrong, because standard form the equation of a parabola is $y = ax^2+bx+c$, there is no cube root in equation of a parabola.

The ratio in which the line segment joining the points $(4, -6)$ and $(3, 1)$ is divided by the parabola $y^2 = 4x$ is

  1. $\displaystyle \frac{-20 \pm \sqrt{155}}{11}: 1$

  2. $\displaystyle \frac{-2 \pm 2\sqrt{155}}{11}: 2$

  3. $-20 \pm 2 \sqrt{155} : 11$

  4. $- 20 \pm \sqrt{155} : 11$


Correct Option: C
Explanation:

Let P(h,k) be a point on the parabola which divides the line segment joining the points A(4,-6) and B(3,1) in the ratio $\lambda:1$.
So, coordinates of point P is $\displaystyle(\frac{3\lambda+4}{\lambda+1},\frac{\lambda-6}{\lambda+1}) $
But this point lies on the parabola
$\displaystyle(\frac{\lambda-6}{\lambda+1})^{2}=4(\frac{3\lambda+4}{\lambda+1})$
$\Rightarrow (\lambda-6)^{2}=4(3\lambda+4)(\lambda+1)$
$\Rightarrow 11\lambda^{2}+40\lambda-20=0$
$\Rightarrow \displaystyle \lambda=\frac{-20\pm 2\sqrt{155}}{11}$
So, the ratio will be ${-20\pm 2\sqrt{155}}:11$

Each member of the family of parabolas $y=ax^2+2x+3$ has a maximum or a minimum point depending upon the value of $a$. The equation of the locus of the maxima or minima for all possible values of $a$ is

  1. a straight line with slope $1$ and $y$ intercept $3$

  2. a straight line with slope $2$ and $y$ intercept $2$

  3. a straight line with slope $1$ and $x$ intercept $3$

  4. a straight line with slope $2$ and $y$ intercept $3$


Correct Option: A
Explanation:
Solution:- (A) a straight line with slope $1$ and $y$-intercept $3$
Consider the general case 
$y = a{x}^{2} + bx + c ..... \left( 1 \right)$. 
Given that the turning point (maximum or minimum) is on the axis of symmetry of the parabola, whose equation is $x = - \cfrac{b}{2a}$, 
$\therefore x$-coordinate of the turning point $= - \cfrac{b}{2a}$.
For $y$-coordinate of the turning point-
Substituting $x = -\cfrac{b}{2a}$ in ${eq}^{n} \left( 1 \right)$, we have

$y = a {\left( - \cfrac{b}{2a} \right)}^{2} + b \left( -\cfrac{b}{2a} \right) + c$
$y = \cfrac{{b}^{2}}{4a} - \cfrac{{b}^{2}}{a} + c$
$\Rightarrow y = -\cfrac{{b}^{2}}{4a} + c$
$\Rightarrow y = \cfrac{bx}{2} + c ..... \left( 2 \right) \; \left[ \text{independent of a} \right]$
Given equation of parabola-
$y = a{x}^{2} + 2x + 3$
Here
$b = 2 \; &amp; \; c = 3$
Substituting these values in ${eq}^{n} \left( 2 \right)$, we get
$y = \cfrac{2x}{2} + 3$
$\Rightarrow y = x + 3$
Hence the equation of the locus of the maxima or minima for all possible values of a is a straight line with slope $1$ and $y$-intercept $3$

The focus of the parabola $y=2x^{2}+x$ is

  1. $(0,0)$

  2. $\left(\dfrac {1}{2},\dfrac {1}{4}\right)$

  3. $\left(-\dfrac {1}{4},\dfrac {1}{8}\right)$

  4. $\left(-\dfrac {1}{4},0\right)$


Correct Option: D
Explanation:
The given equation of parabola is 
$y=2{x}^{2}+x\Rightarrow {x}^{2}+\dfrac{x}{2}=\dfrac{y}{2}$
$\Rightarrow {x}^{2}+2\times \dfrac{x}{2}\times  \dfrac{1}{2}+\dfrac{1}{4}=\dfrac{y}{2}+\dfrac{1}{4}$
$\Rightarrow {\left(x+\dfrac{1}{2}\right)}^{2}=\dfrac{1}{2}\left(y+\dfrac{1}{8}\right)$
is of the form ${X}^{2}=\dfrac{1}{2}Y$ ......$(1)$    where $A=\dfrac{1}{8}$
Focus of $(1)$ is $\left(0,\dfrac{1}{8}\right)$ 
 where $X=0,Y=\dfrac{1}{8}$
$\Rightarrow x+\dfrac{1}{4}=0$ and $y+\dfrac{1}{8}=\dfrac{1}{8}$
$\Rightarrow x=\dfrac{-1}{4}$ and $y=0$
$\therefore$ focus of given parabola is $\left(\dfrac{-1}{4},0\right)$

If the vertex and the focus of a parabola are $\left (-1,1 \right )$ and $\left (2,3 \right )$ respectively, then the equation of the directrix is

  1. $3x+2y+14=0$

  2. $3x+2y-25=0$

  3. $2x-3y+10=0$

  4. none of these


Correct Option: A
Explanation:

slope of axis $= \dfrac{3-1}{2+1} = \dfrac{2}{3}$
Slope of directrix $= \dfrac{-3}{2}$
Vertex is midpoint of foot of directrix and focus thus we get the coordinates oif foot of directrix as $(-4,-1)$
Equation of directrix will be $\dfrac{-3}{2}$  $= \dfrac{y+1}{x+4}$

Option A

The axis of the conic $\displaystyle x^{2}+4y-6x+17=0$ is

  1. $\displaystyle x=5 $

  2. $\displaystyle y=5 $

  3. $\displaystyle x=3 $

  4. $\displaystyle x=-3 $


Correct Option: C
Explanation:

$x^2 + 4y -6x +17 =0$

$\Rightarrow x^2 -6x = -4y-17$
$\Rightarrow x^2 - 6x + 9 = -4y + 8$
$\Rightarrow (x-3)^2 = -4(y-2)$
Let $X = x-3$ and $Y = y-2$, then we get
$X^2 = -4Y$
Comparing the above equation with the standard equation of the parabola, we get that the axis of the parabola is given by $X=0$
Hence, axis of the given parabola will be $x-3 = 0\Rightarrow x =3$
Option $C$ is correct.

The equation of directrix from the following is,

  1. $2x - y = 0$

  2. $x + 2y = 0$

  3. $x + y = 0$

  4. $x + 3y = 0$


Correct Option: B
Explanation:
Let $(a,b)$ be the focus and $y=m{x}$ be directrix 

$\implies \bigg(\dfrac{m{x}-y}{\sqrt{1+m^{2}}}\bigg)^{2}=(x-a)^{2}+(y-b)^{2}$

Differentiating on both sides

$\dfrac{(m{x}-y)(m-\dfrac{d{y}}{d{x}})}{1+m^{2}}=(x-a)+(y-b)\dfrac{d{y}}{d{x}}$

$x$ axis is tangent at $(1,0)$

$\dfrac{m(m)}{1+m^{2}}=1-a\implies a=\dfrac{1}{1+m^{2}}$

$y$ axis  is tangent at $(0,2)$

$\dfrac{-2}{1+m^{2}}=b-2\implies b=\dfrac{2{m}^{2}}{1+m^{2}}$

$(1,0)$ lies on parabola

$\dfrac{(m)^{2}}{1+m^{2}}=(1-a)^{2}+b^{2}$

Substituting $a$ and $b$ values 

$\implies (4{m^2}-1)(m^{2})=0\implies m=0,\pm \dfrac{1}{2}$

For $m=0$ we get $a=1,b=0$ which means that the directrix cuts the parabola which is not possible so $m=\pm \dfrac{1}{2}$

$\implies a=\dfrac{4}{5},b=\dfrac{2}{5}$

So the focus is $\bigg(\dfrac{4}{5},\dfrac{2}{5}\bigg)$

the directrix is $2{y}+x=0$

Hence option $B$ is the answer.

The equation of pair of tangents to a parabola is given by $3x^2 +4y^2 +7xy -2x -y - 5 =0 $ and its focus is (1, 1), then the equation of directrix of the parabola is given by   

  1. 9x - 63y -2 = 0

  2. 59x -63y - 8 = 0

  3. 63x - 59y + 8 = 0

  4. 63x - 9y +2 = 0


Correct Option: A
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