Tag: tangent and normal to an ellipse

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Questions Related to tangent and normal to an ellipse

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 domain of $f(x)=\dfrac 1{\sqrt {x-[x]}}$ is 
  1. $R$
  2. $Z$
  3. $R-Z$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $R-Z$

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)

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

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

  1. x + y = 3

  2. x - y = 3

  3. x + y = 1

  4. x - y = 1
Reveal answer Fill a bubble to check yourself
A Correct answer
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 

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

$y-1=-x+2$

$x+y=3$

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$=-l/m$

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

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

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

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