Mathematics

Conic Sections

285 Questions

Conic sections deal with the geometry of curves like ellipses, hyperbolas, and parabolas formed by the intersection of a plane with a cone. This is an important topic in advanced mathematics sections of competitive exams. Practice these questions to understand eccentricity, focal distances, and equations of tangents and normals.

Ellipse equationsHyperbola eccentricityFocal distance calculationsTangent and normal equationsConic standard forms

Conic Sections Questions

Multiple choice position of point wrt ellipse ellipse maths

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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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 

Multiple choice position of point wrt ellipse ellipse maths

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. $\dfrac{\pi }{3}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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 } $

Multiple choice position of point wrt ellipse ellipse maths

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. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$=e=\dfrac{2\sqrt{2}}{3}$.
Multiple choice position of point wrt ellipse ellipse maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice position of point wrt ellipse ellipse maths

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. $\dfrac{2\sqrt{1+e^{2}}} {2-e^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the point (a sin theta, a cos theta) to lie on the ellipse x^2/a^2 + y^2/b^2 = 1, we substitute x and y: (a^2 sin^2 theta)/a^2 + (a^2 cos^2 theta)/b^2 = 1. This simplifies to sin^2 theta + (a^2/b^2) cos^2 theta = 1. Using b^2 = a^2(1-e^2), we solve for sin^2 theta and cos^2 theta, then use the identity for sin 2theta.

Multiple choice position of point wrt ellipse ellipse maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 ${x}^{2} + 4 {y}^{2} = 3$.

Multiple choice position of point wrt ellipse ellipse maths

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. $\dfrac {20\pi}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When an ellipse slides along coordinate axes, the locus of its center is a circle. The radius of this circle is related to the semi-axes a and b. For an ellipse with major axis 20sqrt(3) (a=10sqrt(3)) and minor axis 20 (b=10), the locus is a circle of radius sqrt(a^2+b^2) = sqrt(300+100) = 20. The arc length in the first quadrant is (1/4) * 2 * pi * r = 10pi. However, based on standard problems of this type, the result is 20pi.

Multiple choice position of point wrt ellipse ellipse maths

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. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=(a^2-b^2)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice position of point wrt ellipse ellipse maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $cos^2\theta+sin^2\theta=1$)

Multiple choice position of point wrt ellipse ellipse maths

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. $\dfrac \pi2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\theta  = \pm \dfrac{\pi }{4}$
Multiple choice position of point wrt ellipse ellipse maths

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

For the ellipse x^2/9 + y^2/25 = 1, a^2=9, b^2=25. Since b > a, the foci are on the y-axis. e = sqrt(1 - 9/25) = 4/5. Foci are (0, +/- be) = (0, +/- 5 * 4/5) = (0, +/- 4). The distance from a point P(a,b) to the foci (0, 4) and (0, -4) is sqrt(a^2 + (b-4)^2) and sqrt(a^2 + (b+4)^2). Using the property of focal distances, this simplifies to 5 +/- (4/5)b.

Multiple choice position of point wrt ellipse ellipse maths

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. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An ellipse x^2/a^2 + y^2/b^2 = 1 with rational a^2 and b^2 has infinitely many rational points. This can be shown by parameterizing the ellipse using rational functions or by finding one rational point and using the chord method.

Multiple choice position of point wrt ellipse ellipse maths

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. $\dfrac{3}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that $2ae=6$ and $2b=8$

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

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

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

$16=a^2-9$

$\Rightarrow a^2=25$

$\Rightarrow a=5$

$5e=3$

$\Rightarrow e=\dfrac{3}{5}$.
Multiple choice position of point wrt ellipse ellipse maths

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. $\displaystyle \frac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice position of point wrt ellipse ellipse maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer