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

Find the equation of normal to the hyperbola $\displaystyle \frac{x^2}{25}\, -\, \displaystyle \frac{y^2}{16}\, =\, 1$ at $(5, 0)$.

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

We know equation of normal to the hyperbola $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$ is given by, $\cfrac{a^2x}{x _1}+\cfrac{b^2y}{y _1}=a^2-b^2$

Thus, required normal is, $5x+\cfrac{16y}{0}=9$
Clearly denominator of second term of L.H.S is $0$ so the equation of line is, $y=0$ 
Hence, option 'A' is correct.  

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

Find the equation of normal to the hyperbola $\displaystyle \frac{x^2}{16}\, -\displaystyle 
\frac{y^2}{9}=1$ at the point $\left ( 6, \displaystyle \frac{3}{2}\sqrt{5}\,\right )$

  1. $8\, \sqrt{5}x\, +\, 18y\, =\, 75\, \sqrt{5}$
  2. $4\, \sqrt{5}x\, +\, 9y\, =\, 25\, \sqrt{5}$
  3. $4\, \sqrt{5}x\, +\, 9y\, =\, 75\, \sqrt{5}$
  4. $8\, \sqrt{5}x\, +\, 18y\, =\, 25\, \sqrt{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Required normal is given by,
$\displaystyle \frac{a^2x}{x _1}+\frac{b^2y}{y _1}=a^2+b^2$
$\Rightarrow \displaystyle \frac{16x}{6}+\frac{9y}{(3\sqrt{5}/2)}=25$
$\Rightarrow 8\, \sqrt{5}x\, +\, 18y\, =\, 75\, \sqrt{5}$

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

The equation of the normal at the positive end of the latusrectum of the hyperbola $x^2-3y^2=144$ is

  1. $\sqrt{3}x+2y=32$
  2. $\sqrt{3}x-3y=48$
  3. $3x+\sqrt{3}y=48$
  4. $3x-\sqrt{3}y=48$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The given hyperbola has the equation $\dfrac{x^2}{12^2}-\dfrac{y^2}{(4\sqrt{3})^2}=1$
Eccentricity of hyperbola = $e = \dfrac{\sqrt{a^2+b^2}}{a}=\dfrac{2}{\sqrt{3}}$
Now, equation of positive latus rectum is $x=ae=8\sqrt{3}$
The end-points of latus rectum are calculated as
$\dfrac{(8\sqrt{3})^2}{12^2}-\dfrac{y^2}{(4\sqrt{3})^2}=1$
$\therefore \dfrac{16}{12}-\dfrac{y^2}{48}=1$
$\therefore y^2=\dfrac{48}{3}$
$\therefore y=\pm 4$
Hence, the positive end is $(8\sqrt{3},4)$.
Now, equation of normal at any point $(x _1,y _1)$ is $\dfrac{a^2x}{x _1}+\dfrac{b^2y}{y _1}=a^2e^2$
$\therefore \dfrac{144x}{8\sqrt{3}}+\dfrac{48y}{4}=48\times 4$
$\therefore 6\sqrt{3}x+12y=48\times 4$
$\therefore \sqrt{3}x+2y=32$
This is the required answer.
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Which one of the following points does not lie on the normal to the hyperbola, $\cfrac { { x }^{ 2 } }{ 16 } -\cfrac { { y }^{ 2 } }{ 9 } =1$ drawn at the point $\left( 8,3\sqrt { 3 }  \right) $?

  1. $\left( 13,-\cfrac { 1 }{ \sqrt { 3 } } \right) $
  2. $\left( 12,\cfrac { 1 }{ \sqrt { 3 } } \right) $
  3. $\left( 11,\sqrt { 3 } \right) $
  4. $\left( 10,\sqrt { 3 } \right) $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$   $\Rightarrow \dfrac{2x}{16} - \dfrac{2y}{9} \dfrac{dy}{dx} = 0$

At $(8, 3\sqrt{3})$
$\dfrac{2\times 8}{16} - \dfrac{2\times 3\sqrt{3}}{9}\dfrac{dy}{dx}=0 \Rightarrow \dfrac{3}{2\sqrt{3}}=\dfrac{dy}{dx}$

Therefore slope of normal $=-\dfrac{1}{\frac{dy}{dx}}$
Equation of normal at $(8, 3\sqrt{3})$, 
$y-3\sqrt{3} = -\dfrac{2}{\sqrt{3}}(x-8)$
Clearly, option (D) does not lies on it.

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

Let $A\left( A\sec { \theta  } ,3\tan { \theta  }  \right) $ and $B\left( A\sec { \phi  } ,3\tan { \phi  }  \right) $ where $\theta +\phi =\cfrac { \pi  }{ 2 } $, be two points on the hyperbola $\cfrac { { x }^{ 2 } }{ 4 } -\cfrac { { y }^{ 2 } }{ 9 } =1$. If $\left( \alpha ,\beta  \right) $ is the point of intersection of normals to the hyperbola at $A$ and $B$, then $\beta=$

  1. $\cfrac { -13 }{ 3 } $
  2. $\cfrac { 13 }{ 3 } $
  3. $\cfrac { 3 }{ 13 } $
  4. $\cfrac { -3 }{ 13 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

equation of hyperbola at point $A(2\sec{\theta} , 3\tan{\theta})$ is

$y+\dfrac{2}{3}\sin{\theta}x = \dfrac{13}{3}\tan{\theta}$   -------  $(i)$

and at point $B(2sec{\phi} , 3\tan{\phi})$ is
$y+\dfrac{2}{3}\sin{\phi}x = \dfrac{13}{3}\tan{\phi}$
now 

putting $\phi = \dfrac{\pi}{2} - \theta$


$y+\dfrac{2}{3}\cos{\theta}x = \dfrac{13}{3}\cot{\theta}$   -----  $(ii)$

now multiplying eq.(i) with  $\cos{\theta}$   and eq (ii) with  $\sin{\theta}$

then subtract both equation we find value of $\beta = -\dfrac{13}{3}$

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

If the sum of the slopes of the normal from a point P to the hyperbola $xy = {c^2}$is equal to $\lambda (\lambda  \in {R^ + })$,then the locus of point P is 

  1. ${x^2} = \lambda {c^2}$
  2. ${y^2} = \lambda {c^2}$
  3. ${xy} = \lambda {c^2}$
  4. ${y^2} = {c^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Equation of rectangular hyperbola is $xy={c}^{2}$

Its rectangular coordinates are $\left(ct,\dfrac{c}{t}\right)$

Equation of normal is $c{t}^{4}-x{t}^{3}+ty-c=0$

Slope$=\dfrac{-coefficient\,of\,x}{coefficient\,of\,y}=\dfrac{{t}^{3}}{t}={t}^{2}$

The normal passes through the point $P\left(h,k\right)$

$\Rightarrow\,c{t}^{4}-h{t}^{3}+tk-c=0$

$\therefore\,$ there exists $4$ roots ${t} _{1},{t} _{2},{t} _{3}$ and ${t} _{4}$

Sum of the roots$={t} _{1}+{t} _{2}+{t} _{3}+{t} _{4}=\dfrac{-coefficient\,of\,{t}^{3}}{coefficient\,of\,{t}^{4}}=\dfrac{-\left(-h\right)}{c}=\dfrac{h}{c}$

Sum of the roots taken two at a time$=\sum{{t} _{i}{t} _{j}}={t} _{1}{t} _{2}+{t} _{2}{t} _{3}+{t} _{3}{t} _{4}+{t} _{4}{t} _{1}+{t} _{2}{t} _{4}+{t} _{1}{t} _{3}=\dfrac{-coefficient\,of\,{t}^{2}}{coefficient\,of\,{t}^{4}}=\dfrac{0}{c}=0$

Now,$\sum{{{t} _{i}}^{2}}=\sum{{\left({t} _{i}\right)}^{2}}$ using ${a}^{2}+{b}^{2}={\left(a+b\right)}^{2}$ for $ab=0$

Sum of squares of slopes of normal from $P$ is
${{t} _{1}}^{2}+{{t} _{2}}^{2}+{{t} _{3}}^{2}+{{t} _{4}}^{2}={\left({t} _{1}+{t} _{2}+{t} _{3}+{t} _{4}\right)}^{2}$

$\Rightarrow\,\lambda={\left(\dfrac{h}{c}\right)}^{2}$

$\Rightarrow\,{h}^{2}=\lambda{c}^{2}$

Replace $h\rightarrow\,x$ we get

${x}^{2}=\lambda{c}^{2}$

$\therefore\,{x}^{2}=\lambda{c}^{2}$ is the required locus at $P$
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

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

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

The 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 ]$
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 a normal of slope $m$ to the parabola ${ y }^{ 2 }=4ax$ touches the hyperbola ${ x }^{ 2 }-{ y }^{ 2 }={ a^2 }$, then

  1. ${ m }^{ 6 }-{ 4m }^{ 4 }-{ 3m }^{ 2 }+1=0$
  2. ${ m }^{ 6 }-{ 4m }^{ 4 }+{ 3m }^{ 2 }-1=0$
  3. ${ m }^{ 6 }+{ 4m }^{ 4 }-{ 3m }^{ 2 }+1=0$
  4. ${ m }^{ 6 }+{ 4m }^{ 4 }+{ 3m }^{ 2 }+1=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation of normal with slope $'m'$ to the parabola $y^2=4ax$ is given by,
$y=mx-2am-am^3$
Also this line touches the hyperbola $x^2-y^2=a^2$
thus using condition of tangency to the hyperbola, $c^2=a^2m^2-b^2$
$(-2am-am^3)^2=a^2(m^2-1)$
$\Rightarrow 4m^2+m^6+4m^4=m^2-1$
$\Rightarrow m^6+4m^4+3m^2+1=0$
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 a normal of slope $m$ to the parabola $y^2 = 4ax$ touches the hyperbola $x^2 - y^2 = a^2$, then

  1. $m^6 - 4m^4 - 3m^2 + 1 =0$
  2. $m^6 - 4m^4 + 3m^2 - 1 = 0$
  3. $m^6 + 4m^4 - 3m^2 + 1 = 0$
  4. $m^6 + 4m^4 + 3m^2 + 1 = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation of normal with slope $'m'$ to the parabola $y^2=4ax$ is given by,
$y = mx-2am-am^3$ (i)

Now given (i) is tangent to the hyperbola $x^2-y^2=a^2$

Thus using condition of tangency, $c^2= a^2m^2-a^2$

$\Rightarrow (2am+am^3)^2=a^2(m^2-1)$

$\Rightarrow (2m+m^3)^2=m^2-1\Rightarrow m^6+4m^4+3m^2+1=0$

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

Let P $(asec \theta,\, btan \theta)$ and Q $(asec \phi,\, btan \phi)$, where $\theta\, +\, \phi\, =\, \displaystyle \frac{\pi}{2}$, be two points on the hyperbola $\displaystyle \frac{x^2}{a^2}\, -\, \frac{y^2}{b^2}\, =\, 1$. If (h, k) is the point of intersection of the normals at P & 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

Normal at $\theta,\, \phi$ are
$\displaystyle \left {

\begin{matrix} ax\, cos\, \theta & + & by\, cot\, \theta\, =\,

a^2\, +\, b^2 \ ax\, cos\, \phi & + & by\, cot\, \phi\, =\,

a^2\, +\, b^2 \end{matrix}\right.$
where $\displaystyle \phi\, =\, \frac{\pi}{2}\, -\, \theta$ and these passes through (h, k).

$\therefore\, ah\, cos \theta\, +\, bk\, cot \theta\, =\, a^2\, +\, b^2$ .....(i)
$ah\, sin \theta\, +\, bk\, tan \theta\, =\, a^2\, +\, b^2$ .....(ii)
Multiply (i) by $sin \theta$ & (ii) by $cos \theta$ & subtract them, 
we get
$\Rightarrow\, (bk\, +\, a^2\, +\, b^2)\, (sin \theta\, -\, cos \theta)\, =\, 0$
$k =-(\cfrac{a^2 + b^2}{b})$
Hence, option 'D' is correct.

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

From any point R two normals which are right angled to one another are drawn to the hyperbola $\displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\left ( a>b \right )$ If the feet of the normals are P and Q then the locus of the circumcentre of the triangle PQR is

  1. $\displaystyle \frac{x^{2}+y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \right )^{2}$
  2. $\displaystyle \frac{x^{2}-y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2}$
  3. $\displaystyle \frac{x^{2}+y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2}$
  4. $\displaystyle \frac{x^{2}+y^{2}}{a^{2}+b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Clearly tangent at P and Q intersect at right-angles at S (say)
$ \displaystyle \Rightarrow $ PSQR is cyclic


$ \displaystyle \Rightarrow $ S lies on director circle of hyperbola
$ \displaystyle \Rightarrow S=\sqrt{a^{2}-b^{2}}\cos \theta , \sqrt{a^{2}-b^{2}}\sin \theta  $

$ \displaystyle \Rightarrow   $ Chord with middle point (h,k) i.e. circumcentre will be same as equation of chord of contact w.r.$ \displaystyle \Rightarrow \perp  $ s

$ \displaystyle \Rightarrow \frac{xh}{a^{2}}-\frac{yk}{b^{2}}=\frac{h^{2}}{a^{2}}-\frac{k^{2}}{b^{2}}$ and $\frac{x\sqrt{a^{2}-b^{2}\cos \theta }}{a^{2}}-\frac{y\sqrt{a^{2}-b^{2}}\cos\theta }{b^{2}}=1 $ are identical comparing and solving we get locus as $ \displaystyle \frac{x^{2}+y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2} $

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

The parabola $( y + 1 ) ^ { 2 } = a ( x - 2 )$ passes through the point $( 1 , - 2 )$ then the equation of its directrix is

  1. $4 x + 1 = 0$
  2. $4 x - 1 = 0$
  3. $4 x + 9 = 0$
  4. $4 x - 9 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of parabola is $(y+1)^2=a(x-2)$


it passes through $(1,-2)$

$\implies (-2+1)^2=a(1-2)\$

$(-1)^2=-a\$

$a=-1$

So the equation of a parabola is 

$(y+1)^2=-1(x-2)\$

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

the directrix of parabola is $x=\dfrac{-1}{4}\$

$4x+1=0$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The exhaustive interval of $\lambda$ for which the equation $\dfrac{x^2}{(\lambda^2-2\lambda-3)}+\dfrac{y^2}{\lambda^2+2\lambda-8}=1$ represents a hyperbola is 

  1. $ \lambda \varepsilon (- \infty, -4) \cup (3, \infty)$
  2. $ \lambda \varepsilon (-4, -1) \cup (2, 3)$
  3. $ \lambda \varepsilon (- \infty, -1) \cup (2, \infty)$
  4. $ \lambda \varepsilon (-4,-1)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Equation of the latus rectum of the hyperbola $(10x - 5)^{2} + (10y - 2)^{2} = 9(3x + 4y - 7)^{2}$ is

  1. $y - 1/5 =-3/4(x - 1/2)$
  2. $x - 1/5 =-3/4(y - 1/2)$
  3. $y + 1/5 =-3/4(x + 1/2)$
  4. $x + 1/5 =-3/4(y + 1/2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is of the form (distance from focus)^2 = e^2 * (distance from directrix)^2. By identifying the focus and directrix, one can determine the latus rectum equation.