Tag: ellipse

Questions Related to ellipse

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

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

  1. $2$

  2. $3$

  3. $1$

  4. $4$

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Correct Option: A

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:

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

  1. $(0, 1)$

  2. $(0, \infty)$

  3. $R$

  4. None of these

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Correct Option: C
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 $m  = \dfrac{a}{b} tan  \theta  \in  R.$

If the length of perpendicular drawn from origin to any normal to the ellipse $\cfrac{{x}^{2}}{16}+\cfrac{{y}^{2}}{25}=1$ is $l$, then $l$ cannot be

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

  1. $4$

  2. $5/2$

  3. $1/2$

  4. $2/3$

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Correct Option: A,B,C,D

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:

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

  1. $e^2=5$

  2. $\displaystyle e^2=\frac{\sqrt{5}+1}{2}$

  3. $\displaystyle e=\frac{\sqrt{5}-1}{2}$

  4. $\displaystyle e^2=\frac{\sqrt{5}-1}{2}$

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Correct Option: D
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)

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

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:

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

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

  2. $e^2-e+1=0$

  3. $e^2+e+1=0$

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

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Correct Option: D
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,  $e^4+e^2-1=0$

The normal to the curve x$^2$ = 4y passing (1,2) is 

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

  1. x + y = 3

  2. x - y = 3

  3. x + y = 1

  4. x - y = 1

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Correct Option: A
Explanation:
Given,

$x^2=4y$

$\Rightarrow 2x=4\dfrac{dy}{dx}$

$\therefore \dfrac{dy}{dx}=\dfrac{x}{2}$

$\dfrac{dy}{dx} _{h,k}=\dfrac{h}{2}$

$\dfrac{-1}{\dfrac{dy}{dx} _{h,k}}=-\dfrac{2}{h}$

Equation of normal

$(y-k)=-\dfrac{2}{h}(x-h)$

Given point, $(1,2)$

$(2-k)=-\dfrac{2}{h}(1-h)$

$k=2+\dfrac{2}{h}(1-h)$

$\therefore k=\dfrac{h^2}{4}$

$\Rightarrow \dfrac{h^2}{4}2+\dfrac{2}{h}(1-h)$

upon solving, we get,

$h=2,k=1$

Hence, the equation of normal is 

$(y-1)=\dfrac{-2}{2}(x-2)$

$y-1=-x+2$

$x+y=3$

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 

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

  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

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Correct Option: B
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$=-l/m$

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

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

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

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

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

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

  1. $\dfrac{\pi}{6}$

  2. $\dfrac{\pi}{3}$

  3. $\dfrac{\pi}{4}$

  4. $\dfrac{\pi}{2}$

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Correct Option: C
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.

Let $L$ be an end of the latus rectum of $y^2 = 4x$. The normal at $L$ meets the curve again at $M$. The normal at $M$ meets the curve again at $N$. The area of $\Delta LMN$ is

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

  1. $\dfrac{1280}{9} sq.$ units

  2. $\dfrac{640}{9} sq.$ units

  3. $\dfrac{320}{9} sq.$ units

  4. $\dfrac{160}{9} sq.$ units

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Correct Option: A

The line $2x+y =3$ cuts the ellipse $4x^2+y^2 =5$ at P and Q . If $\theta$ be the angle between the normals  at these point then $tan \theta$ =

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

  1. $1/2$

  2. $3/4$

  3. $3/5$

  4. $5$

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Correct Option: C