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 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

Length of latusrectum of the ellipse $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{b^{2}}=1$, if the normal, at an end of latusrectum passes through one extremity of the minor axis, then equation of eccentricity of ellipse is

  1. $e^4+e^2-1=0$
  2. <font face="MathJax_Main">$e^3+e^2-1=0$</font>
  3. $e^4+e^2+1=0$

  4. <span class="MathJax_Preview"><span class="MathJax"><span class="math"><font face="MathJax_Main">none of these</font>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a specific derivation problem involving the normal at the end of the latus rectum. The resulting condition for eccentricity e is e^4 + e^2 - 1 = 0.

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)

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