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 maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The equation $x^2y^2-2xy^2-3y^2-4x^2y+8xy+12y=0$ represents.

  1. A pair of lines

  2. Pair of lines and a circle

  3. A pair of lines and a parabola

  4. Four lines forming a square
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Collecting the terms $y^2$ and y the given equation can be written as
$y^2(x^2-2x-3)-4y(x^2-2x-3)=0$
or $(x-3)(x+1)y(y-4)=0$
It represents four lines $x=-1$, $x=3$, $y=0$ and $y=4$.
These two sets of parallel lines form a square of side four.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The locus of a point which moves such that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distances from the other two sides is

  1. Hyperbola

  2. A parabola

  3. An ellipse

  4. A ciircle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
If the triangle is PQR, with $PQ = PR$, take P, Q, R to be the points $(0,b), (–a,0), (a,0)$ respectively.
The equation of the line PQ is 
$y-b=\dfrac{b}{+a}(x)$
$ay-ab=bx$
$bx-ay+ab=0$
and the equation of PR is
$y-b=\dfrac{b}{-a}(x)$
$-ay+ab=bx$
$bx+ay-ab=0$
and the equation of QR is
$y=\dfrac{0}{2a}(x)$
$y=0$
To find the locus of a point $X(h,k)$ which moves so that the square of its distance from QR is equal to the product of its distances from $PQ$ and $PR$. The distance from $X(h,k)$ to PQ is 

$d _{1}=\left | \dfrac{bh-ak+ab}{\sqrt{a^2+b^2}} \right |$
and the distance from $X(h,k)$ to PR is
$d _{2}=\left | \dfrac{bh+ak-ab}{\sqrt{a^2+b^2}} \right |$
The distance from $X(h,k)$ to QR is $|k|$
So according to question 
The distance of $X$ to QR=product of distances from $X$ to PQ and PR 
$k=d _{1}d _{2}$

$k=\dfrac{bh-(ak-ab)}{\sqrt{a^2+b^2}}\times\dfrac{bh+(ak-ab)}{\sqrt{a^2+b^2}}$

$k=\dfrac{b^2h^2-(a^2k^2+a^2b^2-2a^2kb)}{a^2+b^2}$

$a^2k+b^2k=b^2h^2-a^2k^2-a^2b^2+2a^2bk$
Putting $h=x,k=y$
$b^2x^2+(2a^2+b^2)y^2+2a^2by-a^2b^2=0$
Hence above equation represents the pair of straight lines

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the normal at any point $P$ of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ meets the axes in $G$ and $g$ respectively, then $|PG| : |Pg|$ is equal to 

  1. $a:b$
  2. $a^2:b^2$
  3. $b^2:a^2$
  4. $b:a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an ellipse x^2/a^2 + y^2/b^2 = 1, the normal at P(x1, y1) meets the axes at G and g. The ratio PG:Pg is a:b.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

One foot of normal of the ellipse $4x^2$ $+$ 9$y^2$ $= 36 $, that is parallel to the line $2x + y = 3 $, is

  1. $\left ( \dfrac{9}{8}, \dfrac{5}{8} \right )$
  2. $\left ( \dfrac{9}{8}, \dfrac{8}{5} \right )$
  3. $\left ( \dfrac{8}{9}, \dfrac{8}{5} \right )$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The normal to 4x^2 + 9y^2 = 36 (x^2/9 + y^2/4 = 1) parallel to 2x + y = 3 (slope -2). The normal slope is -2. Using the normal equation y = mx - (a^2-b^2)m/sqrt(a^2+b^2m^2), we find the point.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The equation of normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

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

$9x^2+5y^2=45$


Differentiating with respect to $x$, we get,

$18x+10y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{18x}{10y}$

Therefore, Slope of tangent at $(0,3)=\dfrac{-0}{30}=0$

Slope of normal = $\dfrac{1}{0}$

The equation of normal is :
$y-3=\dfrac{1}{0}(x-0)$
$x-0=0$
$x=0$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Find the equation of the normal to the ellipse $9x^2 + 16y^2 = 288$ at the point $(4, 3).$

  1. $4x-3y=7$
  2. $3x-4y=7$
  3. $4x+3y=7$
  4. $3x+4y=7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ellipse may be written as, $\displaystyle \frac{x^2}{32}+\frac{y^2}{18}=1$
$\Rightarrow a^2 = 32, b^2 = 18$
Hence required normal at $(4,3)$ is given by,
$\displaystyle y-3=\frac{3\times 32}{4\times 18}(x-4)\Rightarrow y-3=\frac{4}{3}(x-4)\Rightarrow 4x-3y=7$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The number of normals to the ellipse $\dfrac { { x }^{ 2 } }{ 25 } +\dfrac { { y }^{ 2 } }{ 16 } =1$ which are tangents to the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$ is

  1. 1

  2. 2

  3. 3

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A normal to an ellipse is tangent to a circle if the distance from the center to the normal equals the radius. For the given ellipse and circle, there is only one such normal.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The equation of the normal to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ at the end of latus rectum in quadrant $1^{st}$ and $4^{th}$ is

  1. $x-ey-ae^3=0$
  2. $x+ey-ae^3=0$
  3. $y-ex-be^3=0$
  4. $y+ex-be^3=0$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Ellipse:$\cfrac { { x }^{ 2 } }{ a^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b^{ 2 } } } =$

ends of L.R in 1st and 4th quadrant is $L(ae,\cfrac { b^{ 2 } }{ a } )$ and $L'(ae,\cfrac { -b^{ 2 } }{ a } )$
equation of normal at $L$ and $L'$
at $L$, $\Rightarrow \cfrac { a^{ 2 }x }{ ae } -\cfrac { b^{ 2 }y }{ \cfrac { b^{ 2 } }{ a }  } =a^{ 2 }-b^{ 2 }$
$ \Rightarrow ax-aey=e(a^{ 2 }-b^{ 2 })$
$ \Rightarrow x-ey=\cfrac { a^{ 2 } }{ a } (e^{ 2 })(e)$
$ \Rightarrow x-ey-ae^{ 3 }=0---(i)$
at $L'$ $\Rightarrow \cfrac { a^{ 2 }x }{ ae } -\cfrac { b^{ 2 }y }{ -\cfrac { b^{ 2 } }{ a }  } =a^{ 2 }e^{ 2 }$
$\Rightarrow x+ey-ae^{ 3 }=0---(ii)$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The normal of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at a point $P(x _1,y _1)$ on  it, meets the x-axis in $G$. $PN$ is perpendicular to $OX$, where $O$ is origin. The value of $\frac{l(OG)}{l(ON)}$ is -

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

For an ellipse x^2/a^2 + y^2/b^2 = 1, the foot of the normal on the x-axis gives an x-coordinate related to the center such that l(OG)/l(ON) simplifies to eccentricity squared, e^2, using standard ellipse coordinate geometry properties.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The maximum number of normals that can be drawn from any point outside of an ellipse, in general, is 

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

The equation of normals to an ellipse from a given point leads to a degree 4 polynomial in the eccentric angle parameters, meaning a maximum of 4 normals can be drawn from any point outside an ellipse.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The line $y = mx - \displaystyle \frac{(a^2 - b^2)m }{\sqrt{a^2+ b^2 m^2}}$ is normal to the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for all values of $m$ belongs to:

  1. $(0, 1)$
  2. $(0, \infty)$
  3. $R$

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

The equation of the normal to the given ellipse at the point $P(a  \cos  \theta,  b   \sin  \theta)$ is $ax  \sec  \theta - by  \cos ec \theta = a^2 - b^2$.
$\Rightarrow    \displaystyle y = \left ( \frac{a}{b} \tan  \theta \right)  x - \frac{(a^2 - b^2)}{b} \sin \theta$      (i)
Let      $\displaystyle \frac{a}{b} \tan  \theta = m$, so that
$\displaystyle \sin \theta = \frac{bm}{\sqrt{a^2 + b^2 m^2}}$
Hence, the equation of the normal Equation (i) becomes
$ y = mx - \displaystyle \frac{(a^2 - b^2)m}{\sqrt{a^2 + b^2 m^2}}$
$\therefore    m  \in  R,$ as $m  = \dfrac{a}{b} tan  \theta  \in  R.$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the normal at an end of a latus-rectum of an ellipse $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through one extremity of the minor axis, the eccentricity of the ellipse is given by:

  1. $e^2=5$
  2. $\displaystyle e^2=\frac{\sqrt{5}+1}{2}$
  3. $\displaystyle e=\frac{\sqrt{5}-1}{2}$
  4. $\displaystyle e^2=\frac{\sqrt{5}-1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $a>b$, then one of latus rectum of the ellipse is $(ae, \cfrac{b^2}{a})$

Thus equation of normal at this point is given by,

$\cfrac{a^2x}{ae}-\cfrac{b^2y}{b^2/a}=a^2e^2$
Given it passes through one of minor axis ,which is $(0,-b)$
$\Rightarrow \cfrac{a^2(0)}{ae}-\cfrac{b^2(-b)}{b^2/a}=a^2e^2$
$\Rightarrow e^2=\cfrac{b}{a}$

Now using $e^2=1-\cfrac{b^2}{a^2}$
we get,  $e^4+e^2-1=0$
$e^2=\cfrac{-1+\sqrt{5}}{2}, \cfrac{-1-\sqrt{5}}{2}$(not possible)

$\therefore e^2=\cfrac{-1+\sqrt{5}}{2}$