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 one end of the 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, then:

  1. $e^4-e^2+1=0$
  2. $e^2-e+1=0$
  3. $e^2+e+1=0$
  4. $e^4+e^2-1=0$
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,  $e^4+e^2-1=0$

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

The line $l x + m y = n$ is a normal to the ellipse $\dfrac { x ^ { 2 } } { a ^ { 2 } } + \dfrac { y ^ { 2 } } { b ^ { 2 } } = 1 ,$ if 

  1. $\dfrac { a ^ { 2 } } { l^ { 2 } } + \dfrac { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } { n ^ { 2 } } = \dfrac { b ^ { 2 } } { m ^ { 2 } }$
  2. $\dfrac { a ^ { 2 } } { l ^ { 2 } } + \dfrac { b ^ { 2 } } { m ^ { 2 } } = \dfrac { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } { n ^ { 2 } }$
  3. $\dfrac { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } { n ^ { 2 } } + \dfrac { b ^ { 2 } } { m ^ { 2 } } = \dfrac { a ^ { 2 } } { l ^ { 2 } }$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Normals of slope m to the ellipse are given by

$y=mx\pm \dfrac{m(a^2-b^2)}{\sqrt{a^2+b^2m^2}}$

so for $y=mx+c$

$c=\pm\dfrac{m(a^2-b^2)}{\sqrt{a^2+b^2m^2}}$

$c^2=\dfrac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$

for $lx+my+n=0$

$y=-\dfrac{1}{m}x-\dfrac{n}{m}$

$c=-n/m$

slope$=-l/m$

$c^2=\dfrac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$

$\dfrac{n^2}{m^2}=\dfrac{\dfrac{l^2}{m^2}\left(a^2-b^2\right)^2}{a^2+b^2\dfrac{l^2}{m^2}}$

$\dfrac{a^2m^2+b^2l^2}{l^2m^2}=\dfrac{(a^2-b^2)^2}{n^2}$

$\dfrac{a^2}{l^2}+\dfrac{b^2}{m^2}=\dfrac{(a^2-b^2)^2}{n^2}$.
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The line $5x - 3y = 8\sqrt{2}$ is a normal to the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$. If $\theta$ be the eccentric angle of the foot of this normal , then '$\theta$' is equal to

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{3}$
  3. $\dfrac{\pi}{4}$
  4. $\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
From the equation of ellipse given, we have; $a=5,b=3$

Therefore any point on the ellipse $\left( 5cos\theta , 3sin\theta  \right) .$

Normal at this point to the given ellipse is 
$ 5x sec\theta -3y cosec\theta =25-9=16------(1)$

 Also the equation of given normal is $ 5x-3y=8\sqrt { 2 } -----(2)$

 Also the equation of the normal $(1)$ and $(2)$,

$ \Longrightarrow \dfrac { 5sec\theta  }{ 5 } =\dfrac { 3cosec\theta  }{ 3 } =\dfrac { 16 }{ 8\sqrt { 2 }  }$

$ \Longrightarrow sin\theta =cos\theta =\frac { 1 }{ \sqrt { 2 }  }$

$\Longrightarrow \theta =\frac { \Pi  }{ 4 } $

Option (c) is correct.
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 $\displaystyle x^{2} + 4y^{2} = 16$ at the end of the latus rectum in the first quadrant is

  1. $\displaystyle 2x + \sqrt{3} \left ( y + 3 \right ) = 0$
  2. $\displaystyle 2x = \sqrt{3} \left ( y+ 3 \right )$
  3. $\displaystyle \sqrt{3} x = 2 \left ( y + 3 \right )$

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

Given ellipse may be written as, $\displaystyle \cfrac{x^{2}}{16} + \cfrac{y^{2}}{4} = 1$
$\Rightarrow a^2=16, b^2=4, \therefore e=\sqrt{1-\dfrac14}=\dfrac{\sqrt{3}}2$
Then latus rectum of the ellipse in the first quadrant is $(ae, \cfrac{b^2}{a})\equiv (2\sqrt{3},1)$

Then the point form equation of normal at point $(x _1,y _1)$ is $y-y _1=\displaystyle\frac {y _1a^2}{x _1b^2}(x-x _1), x _1\neq 0$
Thus equation of normal at this point is given by,
$\cfrac{16x}{2\sqrt{3}}-\cfrac{4y}{1}=12$
$\Rightarrow 2x=\sqrt{3}(y+3)$

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

The number of distinct normal lines from the exterior point $\displaystyle \left ( 0, : c \right ), : c > b$ , to the ellipse $\displaystyle \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is

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

Given ellipse is, $\displaystyle \cfrac{x^{2}}{a^2} + \cfrac{y^{2}}{b^2} = 1$
Thus general equation of normal to ellipse with slope $m$ is given by,
$y = mx-\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}$
Given external point through which this line is passing is $P(0,c)$
$\Rightarrow c = -\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}$
$\Rightarrow c^2=\cfrac{(a^2-b^2)^2m^2}{a^2+b^2m^2}$
$\Rightarrow m^2=\cfrac{c^2a^2}{(a^2-b^2)^2-c^2b^2}$


Clearly this is a polynomial of degree two so maximum number of normal that
can be drawn from point $P (0,c)$  to the ellipse is $2$, with slopes $+\infty$ and $-\infty$ corresponding to  two roots of $m.$
But the question asked for distinct normal lines, therefore answer is $1$.

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

Equation of the normal to the ellipse  $4 ( x - 1 ) ^ { 2 } + 9 ( y - 2 ) ^ { 2 } = 36 ,$  which is parallel to the line  $3 x - y = 1 ,$  is

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

The ellipse is (x-1)^2/9 + (y-2)^2/4 = 1. The normal parallel to 3x-y=1 has slope -1/3. Using the normal form y-k = m(x-h) +/- m(a^2-b^2)/sqrt(a^2+b^2m^2), where m is the slope of the normal, we substitute the values to find the equation.

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

If  the equation of normal to the ellipse $\displaystyle  4x^{2}+9y^{2}=36$ at the point $(3, -2)$ is $ px+qy=r$. Find the value of $p+q+r.$

  1. $8$
  2. $9$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of normal is $\dfrac{a^2x}{x _1}-\dfrac{b^2y}{y _1}=a^2-b^2$

$\Rightarrow \dfrac{9x}{3}-\dfrac{4y}{-2}=5$
$\Rightarrow 3x+2y=5$
Therefore, $p+q+r=10$

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

Find the condition that the line  $lx+my=n$ be a normal for ellipse

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

The equation of the normal at $\left( a\cos { \phi ,b\sin { \phi  }  } 

\right) $ to the ellipse $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } }

+\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is
$ax\sec { \phi +by cosec\phi=\left( { a }^{ 2 }-{ b }^{ 2 } \right)  }  \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
and the equation of the line is
$lx+my=n\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Now $(i)$ and $(ii)$ represent the same line
$\therefore

\quad \cfrac { a\sec { \phi  }  }{ l } =\cfrac { b cosec\phi }{ m } =\cfrac {

\left( { a }^{ 2 }-{ b }^{ 2 } \right) }{ n } $
$\Rightarrow

\sin { \phi  } =\cfrac { bn }{ m(a^2-b^2) } \quad \cos { \phi  } =\cfrac { {

\left( { an } \right)  } }{ l(a^2-b^2) } $
Squaring and adding we get
$ \cfrac { { a }^{ 2 } }{ { l }^{ 2 } } +\cfrac { { b }^{ 2 } }{ { m }^{ 2

} } =\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { n

}^{ 2 } } $
Hence. option 'C' is correct.

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

The normal at a point $P$ on the ellipse $x^{2}+4y^{2}=16$ meets the x-axis at $Q.$ If $M$ is the mid point of the line segment $PQ$, then locus of $M$ intersects the latus rectums of the given ellipse at the points.

  1. $\displaystyle \left ( \pm \frac{3\sqrt{5}}{7}, \pm \frac{2}{7} \right )$
  2. $\displaystyle \left ( \pm \frac{3\sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4} \right )$
  3. $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{1}{7} \right )$
  4. $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{4\sqrt{3}}{7} \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given Ellipse $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 4 } =1$

$e=\sqrt{1-\dfrac{b^2}{a^2}} =\dfrac {\sqrt{3}}{2}$

$\because P$ is a point on the ellipse 

So, $P=\left( 4\cos { \theta  } ,2\sin { \theta  }  \right) $

Equation of normal to the ellipse $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 4 } =1$ at point $(x _{1},y _{1})=(4\cos \theta, 2\sin \theta)$ is given by

$a^2 y _1(x-x _1)=b^2 x _1(y-y _1)$ 

$\implies 16\times 2\sin \theta(x-4\cos \theta)=4\times 4\cos \theta(y-2\sin \theta)$

$\implies 2x\sin \theta-8\sin \theta \cos \theta=y\cos \theta-2\sin \theta \cos \theta$

$\implies 2x\sin \theta=y\cos \theta + 6\sin \theta \cos \theta$

$\implies \dfrac{2x}{\cos \theta}=\dfrac{y}{\sin \theta}+6$

$\implies 2x\sec \theta -y\text{cosec} \theta = 6$

It meet the x-axis at $Q(3\cos \theta, 0)$

$\therefore M=\left( \dfrac { 7 }{ 2 } \cos { \theta  } ,\sin { \theta  }  \right) =\left( x,y \right) $

Locus of $M$ is

$\dfrac { { x }^{ 2 } }{ { \left( \dfrac { 7 }{ 2 }  \right)  }^{ 2 } } +\dfrac { { y }^{ 2 } }{ 1 } =1$

Latus rectum of the given ellipse is
$x=\pm ae=\pm \sqrt{16-4}=\pm 2\sqrt{3}$
So locus of $M$ meets the latus rectum at points for which
$\displaystyle y^{2}=1-\frac{12\times 4}{49}=\frac{1}{49}$   $\Rightarrow $   $\displaystyle y=\pm \frac{1}{7}$
Hence, the required point is $\displaystyle \left ( \pm 2\sqrt{3}, \pm \frac{1}{7} \right )$.

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

The eccentric angle of the point where the line, $5x\, -\, 3y\, =\, 8\sqrt{2}$ is a normal to the ellipse $\displaystyle\frac{x^2}{25}\, +\, \frac{y^2}{9}\,=\,1$ is

  1. $\displaystyle\frac{3\pi}{4}$
  2. $\displaystyle\frac{\pi}{4}$
  3. $\displaystyle\frac{\pi}{6}$
  4. $tan^{-1}\,2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of the normal to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ at the point $P(a \cos \theta, b \sin \theta)$ is $ax\sec\theta - bycosec\theta= a^2 - b^2$

Given,ellipse equation as $\dfrac{x^2}{5^2} + \dfrac{y^2}{3^2} = 1$

$\Rightarrow$ Length of major axis, $a=5$ and length of minor axis, $b=3$.

$\therefore$The required equation of normal is $ 5x\sec\theta-3ycosec\theta=5^2-3^2$

$\Rightarrow 5x\sec\theta-3ycosec\theta=16 \dots (1)$

Given normal equation $5x-3y=8\sqrt2$

Multiplying both sides with $\sqrt2$

$\Rightarrow 5\sqrt2 x-3\sqrt2 y=16\dots (2)$

Comparing equation $(1)$ and $(2)$

$\Rightarrow \sec\theta=\sqrt2$

$\Rightarrow \cos\theta=\dfrac{1}{\sqrt2}$

$\Rightarrow \theta =\dfrac{\pi}{4}$

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

On the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } =1$, one of the points at which the normals are parallel to the line $2x-y=1$ is

  1. $\displaystyle \left( \frac { 9 }{ \sqrt { 10 } } ,\frac { 2 }{ \sqrt { 10 } } \right) $
  2. $\displaystyle \left( -\frac { 9 }{ \sqrt { 10 } } ,\frac { 2 }{ \sqrt { 10 } } \right) $
  3. $\displaystyle \left( \frac { 2 }{ \sqrt { 10 } } ,\frac { 9 }{ \sqrt { 10 } } \right) $

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

Given equation of ellipse is $\dfrac {x^2}{4}+\dfrac {y^2}{9}=1$

Let the feet of normal be $P(x,y)$

Normal is parallel to $2x-y=1$

Therefore, slope of normal at $P$ $=2$

and slope of slope of tangent at $P$ $=-\dfrac { 1 }{ 2 } $

Point of contact of tangent in slope from is $\left( \dfrac { \pm { a }^{ 2 }m }{ \sqrt { { a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 } }  } ,\dfrac { { \mp b }^{ 2 } }{ \sqrt { { a }^{ 2 }{ m }^{ 2 }+{ b }^{ 2 } }  }  \right) $

Here $a=2,b=3$ and $m=-\dfrac{1}{2}$

Therefore, the point of contact are:

$\left( \dfrac { \pm \left( 4\times \dfrac { -1 }{ 2 }  \right)  }{ \sqrt { 4\times \dfrac { 1 }{ 4 } +9 }  } ,\dfrac { \mp 9 }{ \sqrt { 4\times \dfrac { 1 }{ 4 } +9 }  }  \right) \\ \left( \dfrac { \mp 2 }{ \sqrt { 10 }  } ,\dfrac { \mp 9 }{ \sqrt { 10 }  }  \right) $

 So, option C is correct.

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 $\displaystyle\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1$ at the positive end of latus rectum is : 

  1. $x\,+\,ey\,+\,e^2a\,=\,0$
  2. $x\,-\,ey\,-\,e^3a\,=\,0$
  3. $x\,-\,ey\,-\,e^2a\,=\,0$

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

$Equation\quad of\quad ellipse:\quad \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\ Co-ordinates\quad of\quad positive\quad latus\quad rectum\quad is:\quad (ae,\frac { { b }^{ 2 } }{ a } )\ Equation\quad of\quad normal\quad at\quad point(x1,y1)\quad is:\ \frac { { a }^{ 2 }x }{ x1 } -\frac { { b }^{ 2 }y }{ y1 } ={ (ae) }^{ 2 }\ \therefore \quad Equation\quad is:\quad \frac { { a }^{ 2 }x }{ ae } -\frac { { b }^{ 2 }y }{ \frac { { b }^{ 2 } }{ a }  } ={ (ae) }^{ 2 }\ Or,\quad \frac { x }{ e } -\frac { y }{ 1 } ={ ae }^{ 2 }\ Or,\quad x-ey-{ ae }^{ 3 }=0$


Option [B]

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

Find the area of the rectangle formed by the perpendiculars from the center of the ellipse $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ to the tangent and normal at a point whose eccentric angle is $\displaystyle\frac{\pi}{4}.$ 

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

The perpendicular distances from the center to the tangent and normal at eccentric angle pi/4 are p1 = ab/sqrt(b^2cos^2(pi/4) + a^2sin^2(pi/4)) and p2 = (a^2-b^2)sin(pi/4)cos(pi/4)/sqrt(a^2sin^2(pi/4) + b^2cos^2(pi/4)). The product of these distances for a rectangle formed by these perpendiculars is the area.

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

Assertion (A): Equation of the normal to the ellipse $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ at $P(\displaystyle \frac{\pi}{4})$ is $5x-3y-8\sqrt{2}=0$
Reason (R): Equation of the normal to the ellipse $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $P(x _{1},y _{1})$ is $\displaystyle \frac{a^{2}x}{x _1}-\frac{b^{2}y}{y _1}=a^{2}-b^2$

  1. Both A and R are true but R is not the correct explanation of A

  2. Both A and R are true and R is the correct explanation of A

  3. A is true but R is false

  4. A is false but R is True
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reason is correct.
Equation of normal in parametric form: $\dfrac{ax}{\cos \theta}-\dfrac{by}{\sin \theta}=a^2-b^2$
$\Rightarrow 5\sqrt 2 x-3\sqrt 2 y=25-9=16$
Therefore, assertion is correct but reason is not the correct explanation 

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

The maximum distance of any normal to the ellipse $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ from the centre is:

  1. $a+b$
  2. $a-b$
  3. $a^{2}+b^{2}$
  4. $a^{2}-b^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of any normal to the ellipse is:
$ \cfrac { ax }{ \cos { \theta  } } -\cfrac { by }{ \sin { \theta  } } =ae$ 
Distance from the center is: $ d=\cfrac { ae }{ \sqrt { \cfrac { { a }^{ 2 } }{ { (\cos { \theta  }) }^{ 2 } } +\cfrac { { b }^{ 2 } }{ { (\sin { \theta  }) }^{ 2 } }  }  }$        -------(1) 
$a,e,b$ are fixed for a given ellipse. So, to maximize $d$ we need to minimize the denominator
$ \therefore E={ a }^{ 2 }({ \sec { \theta  }) }^{ 2 }+{ b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 2 }$
$ \cfrac { DE }{ D\theta  } =2{ a }^{ 2 }({ \sec { \theta  }) }^{ 3 }\tan \theta -{ 2b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 3 }\cot \theta =0$
$ \Rightarrow 2{ a }^{ 2 }({ \sec { \theta  }) }^{ 3 }\tan \theta ={ 2b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 3 }\cot \theta $ 
$\Rightarrow { (\tan \theta ) }^{ 4 }=\cfrac { { b }^{ 2 } }{ { a }^{ 2 } } $ 
$\Rightarrow \tan \theta =\sqrt { \cfrac { b }{ a } } $ 
$\therefore  \sin \theta =\sqrt { \cfrac { b }{ a+b }  }$ and 
$\cos \theta =\sqrt { \cfrac { a }{ a+b }  } $ 
Putting these values in equation 1 we get:
$ d=\cfrac { ae }{ \sqrt { \cfrac { { a }^{ 2 }(a+b) }{ a } +\cfrac { { b }^{ 2 }(a+b) }{ b }  }  } $ 
$\Rightarrow d=\cfrac { ae }{ { (a+b) } } $ 
$\Rightarrow d=\cfrac { ae(a-b) }{ { (a+b)(a-b) } } $ 
$\Rightarrow d=a-b$