Mathematics

Conic Sections

278 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 mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

An ellipse of semi-axis $a,b,$ slides between two perpendicular lines, then the locus of its foci is, (the two lines being taken  as the axes of coordinates)

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

This is a classic locus problem involving an ellipse sliding between two perpendicular axes. The locus of the foci is derived using coordinate geometry.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$ represents an ellipse if

  1. $r>1$
  2. $r>5$
  3. $2 < r< 5$
  4. $r<2$ or $r>5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the equation to represent an ellipse, the denominators must be positive and the coefficients must allow for the standard form x^2/A + y^2/B = 1. This requires 2-r > 0 and r-5 < 0, or vice versa, leading to 2 < r < 5.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Eccentricity of ellipse $\frac{{{x^2}}}{{{a^2} + 1}} + \frac{{{y^2}}}{{{a^2} + 2}} = 1$ is $\frac{1}{{\sqrt 3 }}$  then length of Latusrectum is 

  1. $\frac{8}{{\sqrt 3 }}$
  2. $\frac{4}{{\sqrt 3 }}$
  3. $2\sqrt 3 $
  4. $\frac{{\sqrt 3 }}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\cfrac{x^2}{a^2+1}+\cfrac{y^2}{a^2+2}=1$

Eccentricity of ellipse $=\cfrac{1}{\sqrt 3}$
So, $\cfrac{\sqrt{b^2-a^2}}{a}=\cfrac{1}{\sqrt 3}$
Where ellipse 
$\cfrac{x^2}{a^2+1}+\cfrac{y^2}{a^2+2}=1$
So, here
$\cfrac{\sqrt{a^2+2a^2-1}}{a}=\cfrac{1}{\sqrt 3}$
$\implies a^2=2$
So, equation is 
$\cfrac{x^2}{3}+\cfrac{y^2}{4}=1$
Latus rectum $=\cfrac{2b^2}{a}=\cfrac{2\times 4}{\sqrt 3}=\cfrac{8}{\sqrt 3}$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $\dfrac { x ^ { 2 } } { 10 - a } + \dfrac { y ^ { 2 } } { 4 - a } = 1$ represents an ellipse if

  1. $a < 4$
  2. $a > 4$
  3. $4 < a < 10$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac{x^2}{10-a}+\dfrac{y^2}{4-a}=1$

For this equation to represent an ellipse its eccentricity shoule lie between $0$ and $1$.
$\sqrt{1-\dfrac{b^2}{a^2}}< 1$
$0< 1-\dfrac{(4-a)^2}{(10-a)^2} <1$
$0 < (10-a)^2-(4-a)^2 <1$
$0< 84-12a <1$
$0< (7-a)12<1$
$12(7-a)> 0$ and
$12(7-a)<1$
$a< 7$ and $7-a<\dfrac{1}{12}$
$a< 7$ and $a >\dfrac{83}{12}$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If the latus rectum of an ellipse $x ^ { 2 } \tan ^ { 2 } \varphi + y ^ { 2 } \sec ^ { 2 } \varphi =$ $1$ is $1 / 2 $ then $\varphi $ is

  1. $\pi / 2$
  2. $\pi / 6$
  3. $\pi / 3$
  4. $5$ $\pi/ 12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $x^2 tan^2 \phi + y^2 \, sec^2 \phi = 1$

$\rightarrow \dfrac{x^2}{(1/tan^2 \phi)} + \dfrac{y^2}{(1/sec^2 \phi)} = 1$
$a = \pm \dfrac{1}{tan \phi} , b = \pm \dfrac{1}{sec \phi}$
and $\rightarrow e^2 = 1 - \dfrac{b^2}{a^2}$
$\rightarrow e^2 = 1 - \dfrac{1/sec^2 \phi}{1/tan^2 \phi} = 1 - \dfrac{tan^2 \phi}{sec^2 \phi}$
$\rightarrow e^2 = 1 - sin^2 \phi = cos^2 \phi$
length of latus rectum 
$(LL') = \dfrac{2 b^2}{a} = 2a (1 - e^2)$
$\rightarrow 2a (1 - cos^2 \phi) = 2a. sin^2 \phi = \dfrac{1}{2} $ (Given)
$\therefore 2. \dfrac{cos \phi}{sin \phi} sin^2 \phi = \dfrac{1}{2}$
$\rightarrow 2 cos \phi \, sin \phi = \dfrac{1}{2}$
$\rightarrow sin^2 \phi = \dfrac{1}{2} $
$\rightarrow 2 \phi = \dfrac{\pi}{6} , \dfrac{5 \pi}{6}$
$\therefore \phi = \dfrac{\pi}{12}$    or 
$\phi = \dfrac{ 5 \pi}{12}$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

vertices of an ellipse are $(0,\pm 10)$ and its eccentricity $e=4/5$ then its equation is 

  1. $90x^2-40y^2=3600$
  2. $80x^2+50y^2=4000$
  3. $36x^2+100y^2=3600$
  4. $100x^2+36y^2=3600$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the equation of the required ellipse be 

$\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\longrightarrow \left( 1 \right) $
since the vertices of the ellipse are on $y$-axis, so the coordinate of the vertices are $\left( 0,\pm b \right) $
$\therefore b=10\ Now,\quad { a }^{ 2 }=b^{ 2 }\left( 1-{ e }^{ 2 } \right) \ \Rightarrow { a }^{ 2 }=100\left( 1-\dfrac { 16 }{ 25 }  \right) \ \Rightarrow { a }^{ 2 }=36\ $
substituting the value of ${ a }^{ 2 }$ and ${ b }^{ 2 }$ in equation $(1)$
we get, $\dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 100 } =1\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }=3600\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }-3600=0$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation of the latus rectum of the ellipse $9{x}^{2}+4{y}^{2}-18x-8y-23=0$ are

  1. $y=\pm \sqrt{5}$
  2. $y=- \sqrt{5}$
  3. $y=1\pm \sqrt{5}$
  4. $y=-1\pm \sqrt{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$9x^2+4y^2-18x-8y-23=0$
$=(3x-3)^2+(2y-2)^2-13-23=0$
$\Rightarrow \ 9(x-1)^2+4(y-1)^2=36$
$\Rightarrow \ \dfrac {(x-1)^2}{4}+\dfrac {(y-1)^2}{9}=1$
Shifting origin to $(1,1)\Rightarrow x-1=x,\ y-1=y$
$\Rightarrow \ \dfrac {x^2}{4}+\dfrac {y^2}{9}=1$
$a=2,= b=3,\ e^2=1+\dfrac {a^2}{b^2}=1+\dfrac {4}{9}=\dfrac {5}{9}$
$\Rightarrow \ e=\pm \sqrt {\dfrac {5}{9}}+\dfrac {\sqrt 5}{3}$
$\Rightarrow \ $ Focus $=(0,\ \pm be)=(0,\ \pm \sqrt 5)$
$\Rightarrow \ $ latus ractum $\Rightarrow \ y=\pm \sqrt {5}$
Shifting back, $y=y-1$
$\Rightarrow \ y-1=\pm \sqrt {5}$
$\Rightarrow \ y=1\pm \sqrt 5\  \Rightarrow \ (C) $


Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If there is exactly one tangent at a distance of $4$ units from one of the locus of $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{a^{2}-16}=1, a>4$, then length of latus rectum is :-

  1. $16$
  2. $\dfrac{8}{3}$
  3. $12$
  4. $15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation represents a hyperbola for a > 4. The condition of exactly one tangent at a distance of 4 units implies the distance from the center to the tangent is 4, which corresponds to the distance to the asymptotes or a specific property of the hyperbola. Solving for the latus rectum length 2(a^2-16)/a leads to 16.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The equation $\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$ represents an ellipse, if

  1. $r>2$
  2. $r\in \left(2,\:\dfrac{7}{2}\right)\cup \left(\dfrac{7}{2},5\right)$
  3. $r>5$
  4. $r<2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Equating the equation of the ellipse with the second-degree equation
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ with $\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$
we get $A=\dfrac{1}{2-r}, B=0, C=\dfrac{1}{r-5},D=0,E=0$ and $F=1$
For the second degree equation to represent an ellipse, the coefficients must satisfy the discriminant condition ${B}^{2}-4AC<0$ and also $A\neq C$
$\Rightarrow {\left(0\right)}^{2}-4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$
$\Rightarrow -4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$
$\Rightarrow \left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)>0$
$\Rightarrow \left(2-r\right)\left(r-5\right)<0$
$\Rightarrow \left(r-2\right)\left(r-5\right)>0$
$\Rightarrow r>2$
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Equation of the ellipse whose minor axis is equal to the distance between foci and whose latus rectum is $10 ,$ is given by ____________.

  1. $2 x ^ { 2 } + 3 y ^ { 2 } = 100$
  2. $2 x ^ { 2 } + 3 y ^ { 2 } = 80$
  3. $x ^ { 2 } + 2 y ^ { 2 } = 100$

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

Let the ellipse be x^2/a^2 + y^2/b^2 = 1. Minor axis 2b = distance between foci 2ae, so b = ae. Latus rectum 2b^2/a = 10. Using b^2 = a^2(1-e^2) = a^2 - a^2e^2 = a^2 - b^2, we get a^2 = 2b^2. Substituting into the latus rectum formula: 2b^2/sqrt(2b^2) = 10, so sqrt(2)b = 10, b^2 = 50, a^2 = 100. The equation is x^2/100 + y^2/50 = 1, which multiplies to x^2 + 2y^2 = 100.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

For the ellipse $ {12x}^{2} +{4y}^{2} +24x-16y+25=0 $

  1. centre is $(-1,2) $
  2. Length of axes are $ {\sqrt {3}} and 1 $
  3. eceentricity is $ \sqrt {\cfrac {2} {3}} $

  4. All of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$12x^2+4y^2+24x-16y+25=0$

$\Rightarrow 12(x+1)^2+4(y-2)^2=3$

$\dfrac{(x+1)^2}{\frac{1}{4}}+\dfrac{(y-2)^2}{\frac{3}{4}}=1$

$\therefore a=\dfrac{1}{2},b=\dfrac{\sqrt 3}{2}$

⇒ Centre $ = (-1, 2)$

Here $b^2>a^2$

⇒ eccentricity$(e) = \sqrt {\dfrac {b^2-a^2}{b^2}} $

$= \sqrt {\dfrac {\dfrac 3 4 - \dfrac 1 4}{\dfrac 3 4}}=\sqrt{\dfrac 2 3}$

Length of arcs,

length of major arc $=2b=2\left ( \dfrac{\sqrt 3}{2} \right )=\sqrt 3$

length of minor arc $=2a=2\left ( \dfrac{1}{2} \right )=1$

Option D is correct.
Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

A point $P$ on the ellipse $\displaystyle \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ has the eccentric angle $\displaystyle \frac{\pi}{8}$. The sum of the distance of $P$ from the two foci is

  1. $5$
  2. $6$
  3. $10$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,ellipse equation as $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Length of major axis, $a=5$ and length of minor axis, $b=3$
$P$ is a point on the ellipse whose eccentricity is $\dfrac{\pi}{8}.$
We know that, sum of the distances of any point on the ellipse from its foci equal to twice the major axis.
Let $S,S'$ be foci of ellipse and $a,b$ as the length of major,minor axis respectively.
$\Rightarrow SP+S'P=2a$
$\Rightarrow SP+S'P=2 \times 5=10$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Axes are coordinates axes, the ellipse passes through the points where the straight line $\dfrac {x}{4}+\dfrac {y}{3}=1$  meets the coordinates axes. Then equation of the ellipses is 

  1. $\dfrac {x^{2}}{16}+\dfrac {y^{2}}{9}=1$
  2. $\dfrac {x^{2}}{64}+\dfrac {y^{2}}{36}=1$
  3. $\dfrac {x^{2}}{4}+\dfrac {y^{2}}{3}=1$
  4. $\dfrac {x^{2}}{8}+\dfrac {y^{2}}{6}=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line x/4 + y/3 = 1 meets the x-axis at (4, 0) and the y-axis at (0, 3). An ellipse centered at the origin passing through these points has the form x^2/a^2 + y^2/b^2 = 1. Substituting the points gives a^2 = 16 and b^2 = 9.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If $(2,4)$ and $( 10,10)$ are the ends of a latus - rectum of an ellipse with eccentricity $\dfrac 12$, then the length of semi - major axis is 

  1. $\dfrac{20}{3}$
  2. $\dfrac {15}{3}$
  3. $\dfrac {40}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $(2,4)$ and $(10,10)$ are the ends of the latusrectum and eccentricity is $\dfrac{1}{2}$

We know that length of the latus rectum is $\dfrac{2b^{2}}{a}$

We know that the distance between the two points

$(x _{1}, y _{1})$ and $(x _{2}, y _{2})$ is

$\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})^{2}}$

$\Rightarrow \dfrac{2b^{2}}{a^{2}}=\sqrt{(2-10)^{2}+(4-10)^{2}}$

$\Rightarrow \dfrac{2b^{2}}{a}=\sqrt{(-8)^{2}+(-6)^{2}}=\sqrt{64+36}=\sqrt{100}$

$\Rightarrow 36^{2}=10a$

$\Rightarrow b^{2}=5a$

we know that $b^{2}=a^{2}(1-e^{2})$

$a^{2}(1-e^{2})=5a$

$a\left(1-\left(\dfrac{1}{2}\right)^{2}\right)=5$

$a=\dfrac{5}{1-\dfrac{1}{4}}=\dfrac{5}{\dfrac{3}{4}}$

$a=\dfrac{20}{3}$

Thus the length of semi major axis is $\dfrac{20}{3}$