Tag: normal to an ellipse

Questions Related to normal to an ellipse

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

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

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

  4. none of these

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Correct Option: A
The domain of $f(x)=\dfrac 1{\sqrt {x-[x]}}$ is 
maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $R$

  2. $Z$

  3. $R-Z$

  4. $None\ of\ these$

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Correct Option: C
Explanation:

The function is defined as 

$f(x)=\dfrac 1{\sqrt {x-[x]}}$

The function is not defined if

$\sqrt {x-[x]}=0$

$\implies x-[x]=0$

$\implies x=[x]$

This happens only in the case of Integers 

So The function is not defined at Integer Values

Hence , The Domain of function is $R-Z$

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 -

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

  1. $e$

  2. $e^2$

  3. $e^3$

  4. $e^2-1$

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Correct Option: C

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$=-l/m$

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

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

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

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