Mathematics · Quantitative Aptitude

Conic Sections

239 Questions

Conic sections deal with the curves obtained by the intersection of a cone with a plane, primarily focusing on parabolas, ellipses, and hyperbolas. Questions require finding vertices, directrices, and asymptotes based on given equations. This is an advanced geometry topic for rigorous competitive exams.

Parabola equationsHyperbola propertiesEllipse conceptsTangents and normalsDirectrix and focus

Conic Sections Questions

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The normal to the hyperbola $4x^2-9y^2=36$ meets the axes in $M$ and $N$ and the lines $MP$, $NP$ are drawn right angles at the axes. The locus of $P$ is the hyperbola 

  1. $9x^2-4y^2=169$
  2. $4x^2-9y^2=169$
  3. $3x^2-4y^2=169$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac {x^2}9-\dfrac {y^2}4=1$.Let $P(x _1, y _1)$ be the point on hyperbola.

Eqn of the normal is$\dfrac {a^2x}{x _1}-\dfrac {b^2y}{y _1}=a^2b^2\M=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { x } _{ 1 }=x\N=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { y } _{ 1 }=y\P=(x, y)$$x _1=\dfrac {a^2(x)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.$y _1=\dfrac {b^2(y)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.Now, $a^2=9, b^2=4$Therefore, $x _1=\dfrac {9x}5, y _1=\dfrac {4y}5$$\dfrac {x _1^2}9-\dfrac {y _1^2}4=1\\left(\dfrac {9x}5\right)^2\dfrac {x _1^2}9-\left(\dfrac {4y}5\right)^2\dfrac {y _1^2}4=1\\implies 9x^2-4y^2=25$

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

A normal to the hyperbola, $4x^2-9y^2=36$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram $OABP$($O$ being the origin) is formed, then the locus of $P$ is?

  1. $4x^{2}+9y^{2}=121$
  2. $9x^{2}+4y^{2}=169$
  3. $4x^{2}-9y^{2}=121$
  4. $9x^{2}-4y^{2}=169$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the hyperbola 4x^2 - 9y^2 = 36, the normal equation and the locus of the vertex P of the parallelogram OABP result in 9x^2 - 4y^2 = 169.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Equation of the normal to the hyperbola $3x^2-y^2=3$ at $(2, -3)$ is?

  1. $x-2y-8=0$
  2. $3x-2y-12=0$
  3. $x+2y+4=0$
  4. $3x+2y-14=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the hyperbola 3x^2 - y^2 = 3, the derivative at (2, -3) gives the slope of the tangent. The normal slope is the negative reciprocal. Using the point-slope form, the equation is x - 2y - 8 = 0.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Line x cos$\alpha $+yin$\alpha $=p is a normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $, if

  1. $a^{2}sec^{2}\alpha -b^{2}cosec^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
  2. $a^{2}sec^{2}\alpha+b^{2}cosec^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
  3. $a^{2}cos^{2}\alpha -b^{2}sin^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
  4. $a^{2}cos^{2}\alpha+b^{2}sin^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The condition for the line x cos(alpha) + y sin(alpha) = p to be a normal to the hyperbola x^2/a^2 - y^2/b^2 = 1 is a^2 sec^2(alpha) - b^2 csc^2(alpha) = (a^2 + b^2)^2 / p^2.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Line $ x \cos \alpha + y \sin \alpha = p $ is a normal to the hyperbola $ \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 $, if 

  1. $ a ^ { 2 } \sec ^ { 2 } \alpha - b ^ { 2 } \csc ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
  2. $ a ^ { 2 } \sec ^ { 2 } x + b ^ { 2 } \csc ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
  3. $ a ^ { 2 } \cos ^ { 2 } \alpha - b ^ { 2 } \sin ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
  4. $ a ^ { 2 } \cos ^ { 2 } \alpha + b ^ { 2 } \sin ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is identical to the previous question (424646). The condition for the line to be a normal is a^2 sec^2(alpha) - b^2 csc^2(alpha) = (a^2 + b^2)^2 / p^2.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

A straight line is drawn parallel to the conjugate axis of the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ to meet it and the conjugate hyperbola respectively in the point $P$ and $Q$. The normals at $p$ and $Q$ to the curves meet on 

  1. $x-axis$
  2. $y-axis$
  3. $y=x$
  4. $y=-x$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Equation of hyperbola $\to \dfrac {x^2}{9^2}-\dfrac {y^2}{6^2}=1$
Conjugate hyperbola $\to \dfrac {y^2}{6^2}-\dfrac {x^2}{9^2}=1$
Line $||$ to conjugate axis of hyperbola $(y-axis)$ is drawn to meet conjugate axis at $P$ & $Q$ which are symmetric points about $x-$axis then
$P(a\tan \theta , b\sin theta)$
$Q(a \tan \theta , -b\sin \theta)$
Normal at $P\to \dfrac {ax}{\tan \theta} +\dfrac {6y}{\sec \theta}=a^2+b^2$
Normal at $Q\to \dfrac {ax}{\tan \theta}-\dfrac {6y}{\sec \theta}=a^2+b^2$
Let us find out interrection
$\dfrac {b}{\sec \theta}=\dfrac {-6y}{\sec \theta} $
$y=0$
$9+$ lies on $x-$axis
$A$ is correct


Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the normal at $\left (ct _1,\dfrac { c}{t _1}\right)$ on the hyperbola $xy = c^2$ cuts the hyperbola again at $\left (ct _2, \dfrac {c}{t _2}\right)$, then $t _2^3 t _2$ $=$ 

  1. $2$
  2. $-2$
  3. $-1$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of hyperbola is $xy=c^2$ and point $(ct _1,\dfrac{c}{t _1})$ lies on it.


Let us find the equation of the normal.


Equation of hyprbola can be written as $y=\dfrac{c^2}{x}$ and therefore slope of tangent is given by first derivative i.e. $\dfrac{dy}{dx}=−\dfrac{c^2}{x^2}$


hence slope of normal is given by $\dfrac{x^2}{c^2}$ and at $(ct _1,\dfrac{c}{t _1})$ is $t^2$ and its equation is


$y=t _1^2(x−ct _1)+\dfrac{c}{t _1}$


or $xt _1^3−yt _1−ct _1^4+c=0$


As this passes through $(ct _2,\dfrac{c}{t _2})$


$ct _2t _1^3−\dfrac{c}{t _2}t _1−ct _1^4+c=0$


or $ct _1^3(t _2−t _1)+\dfrac{c}{t _2}(t _2−t _1)=0$


as $t _1\neq t _2, t _1−t _2\neq 0$ and dividing by it we get


$ct _2^3=−\dfrac{c}{t _2}$


Or $ t _2^3t _2=−1$

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the tangent and normal to a rectangular hyperbola cut off intercepts $x _1$ and $x _2$ on one axis and $y _1$ and $y _2$ on the other axis, then

  1. $x _1y _1+x _2y _2=0$
  2. $x _1y _2+x _2y _1=0$
  3. $x _1x _2+y _1y _2=0$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Assume rectangular hyperbola is $xy = c^2$
Thus equation of tangent and normal at any point 't' are,
$\cfrac{x}{t}+ty=2c$ and $ y-\cfrac{c}{t}=t^2(x-ct)$
Now putting $y=0$ in both the equation we get, $x _1=2ct, x _2=ct-\cfrac{c}{t^3}$
and putting $x=0$ we get, $y _1=\cfrac{2c}{t}, y _2=\cfrac{c}{t}-ct^3$
$\Rightarrow x _1x _2+y _1y _2=2ct(ct-\cfrac{c}{t^3})+\cfrac{2c}{t}(\cfrac{c}{t}-ct^3)=0$
Hence, option 'C' is correct.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The number of normal to the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ from an external point is

  1. $2$
  2. $4$
  3. $6$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given hyperbola is, $\displaystyle \cfrac{x^{2}}{a^2} - \cfrac{y^{2}}{b^2} = 1$
The general equation of normal to hyperbola with slope 'm' is given by,
$y = mx\pm\cfrac{(a^2+b^2)m}{\sqrt{a^2-b^2m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{(a^2+b^2)^2m^2}{a^2-b^2m^2}$
$\Rightarrow (y _1-mx _1)^2(a^2-b^2m^2)=(a^2+b^2)^2m^2$
Clearly, this is a polynomial of degree four so maximum number of normal that can be drawn from point P (any external point) to the hyperbola is 4 corresponding to four roots of m.
Hence, option 'B' is correct.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The normal to the rectangular hyperbola $xy=-c^2$ at the point $'t _1'$ meets the curve again at the point $'t _2'$. The value of $t _1^3 \cdot t _2$ is

  1. $1$
  2. $c$
  3. $-c$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of the normal $t _{1}$is  $y-\dfrac{c}{t _{1}}=t _{1}^{2}(x-ct _{1})$

If this passes through $\left ( ct _{2},\dfrac{c}{t _{2}} \right )$

$\dfrac{\mathrm{c}}{\mathrm{t} _{2}}-\dfrac{\mathrm{c}}{\mathrm{t} _{\mathrm{t}}}=\mathrm{t} _{1}^{2}(\mathrm{c}\mathrm{t} _{2}-\mathrm{c}\mathrm{t} _{1})$

$\displaystyle



\Rightarrow-\dfrac{1}{\mathrm{t} _{\mathrm{t}}\mathrm{t} _{2}}=\mathrm{t} _{1}^{2}\Rightarrow

1+\mathrm{t} _{\mathrm{t}}^{3}\mathrm{t} _{2}=0$

$\Rightarrow t _1^3t _2=-1$

Hence, option 'D' is correct.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the normal at '$\theta $' on the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ meets the transverse axis at $G$  and $A$ and $A'$ are the vertices of the hyperbola, then $AG.A'G$ $=$

  1. ${ a }^{ 2 }\left( { e }^{ 2 }\sec ^{ 2 }{ \theta } -1 \right) $
  2. ${ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta } -1 \right) $
  3. ${ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta } +1 \right) $

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of the normal at $\left( a\sec { \theta  } ,b\tan { \theta  }  \right) $ to the given hyperbola is
$ax\cos { \theta  } +by\cot { \theta  } =\left( { a }^{ 2 }+{ b }^{ 2 } \right) $
This meets the transverse axis (i.e) at $G$. So, the coordinates of $G$ are $\left{ \cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  }
,0 \right} $
The coordinates of the vertices $A$ and $A'$ are $A(a,0)$ and $A'(-a,0)$ respectively
$\therefore

\quad AG.A'G=\left( -a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec {

\theta  }  \right) \left( a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec

{ \theta  }  \right)$
 $\Rightarrow AG.A'G=\left( -a+a{ e }^{ 2 }\sec { \theta  }  \right) \left( a+a{ e }^{ 2 }\sec { \theta  }  \right) $
$={ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta  } -1 \right) $
Hence, option 'B' is correct.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the normal at $'\theta'$ on the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ meets the transverse axis at G, and A and A' are the vertices of the hyperbola, then AG.A'G $=$

  1. $a^2 (e^2 sec^2 \theta -1)$
  2. $a^2 (e^4 sec^2 \theta - 1)$
  3. $a^2 (e^4 sec^2 \theta + 1)$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of the normal at $\left( a\sec { \theta  } ,b\tan { \theta  }  \right) $ to the given hyperbola is
$ax\cos { \theta  } +by\cot { \theta  } =\left( { a }^{ 2 }+{ b }^{ 2 } \right) $
This meets the transverse axis (i.e) at $G$. So, the coordinates of $G$ are $\left{ \cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  }

,0 \right} $
The coordinates of the vertices $A$ and $A'$ are $A(a,0)$ and $A'(-a,0)$ respectively
$\therefore

\quad AG.A'G=\left( -a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec {

\theta  }  \right) \left( a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec

{ \theta  }  \right)$
 $\Rightarrow AG.A'G=\left( -a+a{ e }^{ 2 }\sec { \theta  }  \right) \left( a+a{ e }^{ 2 }\sec { \theta  }  \right) $
$={ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta  } -1 \right) $
Hence, option 'B' is correct.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The equation of normal at $\left( at,\dfrac { a }{ t }  \right)$ to the hyperbola $xy={ a }^{ 2 }$ is ________________________.

  1. ${ xt }^{ 3 }-yt+{ at }^{ 4 }-a=0$
  2. ${ xt }^{ 3 }-yt-{ at }^{ 4 }+a=0$
  3. ${ xt }^{ 3 }+yt+{ at }^{ 4 }-a=0$
  4. ${ xt }^{ 3 }+yt-{ at }^{ 4 }-a=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the rectangular hyperbola xy = a^2, the normal at (at, a/t) is given by the equation xt^3 - yt - at^4 + a = 0.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The maximum number of normals to the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2}=1$ from an external point is :

  1. $2$
  2. $4$
  3. $6$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given hyperbola is, $\displaystyle \cfrac{x^{2}}{a^2} - \cfrac{y^{2}}{b^2} = 1$
The general equation of normal to hyperbola with slope $m$ is given by,
$y = mx\pm\cfrac{(a^2+b^2)m}{\sqrt{a^2-b^2m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{(a^2+b^2)^2m^2}{a^2-b^2m^2}$
$\Rightarrow (y _1-mx _1)^2(a^2-b^2m^2)=(a^2+b^2)^2m^2$
Clearly, this is a polynomial of degree four so maximum number of normal that can be drawn from point $P$ (any external point) to the hyperbola is $4$ corresponding to four roots of $m$.
Hence, option 'B' is correct.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Let $P (a\sec \theta , b\tan \theta ) $ and $Q\left ( a\sec \phi , b\tan \phi  \right )$ where $\theta +\phi =\pi /2$, be two points on the hyperbola $x^{2}/a _{2}-y _{2}/b _{2}=1$. If (h, k) is the point of intersection of normals at P and Q, then k is equal to

  1. $\displaystyle \frac{a^{2}+b^{2}}{a}$
  2. $\displaystyle -\left [ \frac{a^{2}+b^{2}}{a} \right ]$
  3. $\displaystyle \frac{a^{2}+b^{2}}{b}$
  4. $\displaystyle -\left [ \frac{a^{2}+b^{2}}{b} \right ]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equations of the normal at P is $ax+bycosec\theta =\left ( a^{2}+b^{2} \right )\sec \theta $          (i)
and the equation of the normal at $Q\left ( a\sec \phi , b\sec \phi  \right )$ is
$ax+by cosec\phi =\left ( a^{2}+b^{2} \right )\sec \phi $          (ii)
Subtracting (ii) from (i) we get

   $\displaystyle y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec \phi }{cosec \theta -cosec \phi }$

So $\displaystyle k=y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec

\left ( \pi /2-\theta  \right )}{cosec \theta -cosec \left ( \pi

/2-\theta  \right )}$          $\left [ \because \theta +\phi =\pi /2

\right ]$

     $\displaystyle =\frac{a^{2}+b^{2}}{b}.\frac{\sec

\theta -cosec \theta }{cosec \theta -\sec \theta }=-\left [

\frac{a^{2}+b^{2}}{b} \right ]$