Tag: normal to an ellipse

Questions Related to normal to an ellipse

The maximum distance of the normal to the ellipse $\displaystyle \frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$ from its centre is:

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

  1. $\displaystyle \frac{1}{2}$

  2. $2$

  3. $1$

  4. $4$

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Correct Option: C
Explanation:
Ellipse : $\cfrac { { x }^{ 2 } }{ 9 } +\cfrac { { y }^{ 2 } }{ 4 } =1$

Equation of the normal,
$\cfrac { { ax }^{  } }{ \cos  { \theta  }  } -\cfrac { { by }^{  } }{ \sin { \theta  } } ={ a }^{ 2 }-{ b }^{ 2 } \\ \therefore \cfrac { { 3x }^{  } }{ \cos  { \theta  }  } -\cfrac { { 2y }^{  } }{ \sin { \theta  } } =5$

Or, $\ { 3x }^{  }\sin { \theta  }-{ 2y }^{  }\cos  { \theta  } =5\cos  { \theta  } \sin { \theta  }$

Distance from origin d= $\cfrac { \left| 0+0-5\cos  { \theta  } \sin { \theta  } \right|  }{ \sqrt { 9{ \left( \cos  { \theta  }  \right)  }^{ 2 }+{ 4\left( \sin { \theta  } \right)  }^{ 2 } }  } $

Or, d=$\cfrac { 5 }{ \sqrt { 9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 } }  } $

To maximize d we need to minimize the denominator.
$E=9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 }\ then,\quad \\\cfrac { dE }{ d\theta  } =-18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \ For\quad \\Minimizing,\quad \cfrac { dE }{ d\theta  } =0\\ \therefore -18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } =0 \\Or,18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } =8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \\ Or,{ \left( \tan { \theta  }  \right)  }^{ 4 }=\cfrac { 9 }{ 4 } \\ Or,\quad \tan { \theta  } =\sqrt { \cfrac { 3 }{ 2 }  } \\ \therefore \csc { \theta  } =\sqrt { \cfrac { 5 }{ 3 }  } \quad \quad and\quad \quad \sec { \theta  } =\sqrt { \cfrac { 5 }{ 2 }  } $

 On putting the values in d we get,
$d=\cfrac { 5 }{ \sqrt { 15+10 }  } \\ Or,\quad d=1$

lf the tangent drawn at a point $(t^{2},2t)$ on the parabola $y^{2}=4x$ is same as normal drawn at $(\sqrt{5}\cos\alpha, 2\sin\alpha)$ on the ellipse $\displaystyle \frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, then which of following is not true?  

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

  1. $t=\displaystyle \pm\frac{1}{\sqrt{5}}$

  2. $\alpha=-\tan^{-1}2$

  3. $\alpha=\tan^{-1}2$

  4. $\alpha=\tan^{-1}4$

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Correct Option: D

If the line $x\cos { \alpha  } +y\sin { \alpha  } =p$ be normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, then

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

  1. ${ p }^{ 2 }\left( { a }^{ 2 }\cos ^{ 2 }{ \alpha } +{ b }^{ 2 }\sin ^{ 2 }{ \alpha } \right) ={ a }^{ 2 }-{ b }^{ 2 }$

  2. ${ p }^{ 2 }\left( { a }^{ 2 }\cos ^{ 2 }{ \alpha } +{ b }^{ 2 }\sin ^{ 2 }{ \alpha } \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right) }^{ 2 }$

  3. ${ p }^{ 2 }\left( { a }^{ 2 }\sec ^{ 2 }{ \alpha } +{ b }^{ 2 }\csc ^{ 2 }{ \alpha } \right) ={ a }^{ 2 }-{ b }^{ 2 }$

  4. ${ p }^{ 2 }\left( { a }^{ 2 }\sec ^{ 2 }{ \alpha } +{ b }^{ 2 }\csc ^{ 2 }{ \alpha } \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right) }^{ 2 }$

Show answer
Correct Option: D
Explanation:

The equation of any normal to $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is 
       $ax\sec { \phi  } -by\csc { \phi  } ={ a }^{ 2 }-{ b }^{ 2 }$              ......(i)
The straight line $x\cos { \alpha  } +y\sin { \alpha  } =p$ will be a normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, if equation (i) and $x\cos { \alpha  } +y\sin { \alpha  } =p$ represent the same line.
$\therefore \dfrac { a\sec { \phi  }  }{ \cos { \alpha  }  } =\dfrac { -b\csc { \phi  }  }{ \sin { \alpha  }  } =\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ p } $
$\Rightarrow \cos { \phi  } =\dfrac { ap }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \cos { \alpha  }  } $
$\sin { \phi  } =\dfrac { -bp }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \sin { \alpha  }  } $
$\because \sin ^{ 2 }{ \phi  } +\cos ^{ 2 }{ \phi  } =1$
$\Rightarrow \dfrac { { b }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\sin ^{ 2 }{ \alpha  }  } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\cos ^{ 2 }{ \alpha  }  } =1$
$\Rightarrow { p }^{ 2 }\left( { b }^{ 2 }\csc ^{ 2 }{ \alpha  } +{ a }^{ 2 }\sec ^{ 2 }{ \alpha  }  \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }$

If the line $x \cos a + y \sin a = p$ be normal to the ellipse $\dfrac{x^2}{a^2}$ $+\dfrac{y^2}{b^2}$ = 1 then

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

  1. $p^2(a^2\cos^2a+b^2\sin^2a)=a^2-b^2$

  2. $p^2(a^2\cos^2a+b^2\sin^2a)=(a^2-b^2)^2$

  3. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)$

  4. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$

Show answer
Correct Option: A
Explanation:

A line $y=mx+c$ is normal to ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if $c^2=m^2\dfrac{(a^2-b^2)^2}{a^2+m^2b^2}$
Given equation $x\cos a+y\sin a=p\Rightarrow y=-(\dfrac{\cos a}{\sin a})x+\dfrac{p}{\sin a}$
Here $m=-\cot a, c=\dfrac{p}{\sin a}$
Substituting in the formulae, we get
$\dfrac{p^2}{\sin ^2a}=\dfrac{\cos ^2a}{sin^2a}\times \dfrac{(a^2-b^2)^2}{a^2+(\cot^2a) b^2}$
After simplification, we get
$p^2\dfrac{(a^2\sin^2a+b^2\cos^2a)}{\sin^2a\cos^2a}=(a^2-b^2)^2$
$p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$

If the normal at the point $P(\theta)$ to the ellipse $\dfrac {x^{2}}{14} + \dfrac {y^{2}}{5} = 1$ intersects it again at the point $Q(2\theta)$, then $\cos \theta$ is equal to

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

  1. $2/3$

  2. $-2/3$

  3. $3/4$

  4. None of these

Show answer
Correct Option: B
Explanation:
Normal at the point P(theta) to the ellipse $\dfrac{x²}{14} + \dfrac{y²}{5 }= 1$ intersects it again at the point $Q(2\  \theta). $

we know, standard equation of ellipse is 

$\dfrac{x²}{a²} + \dfrac{y²}{b²} = 1 $ compare it with given equation

so, $ a² = 14$ then, $a = √14 $

$b² = 5$ then, $b = √5 $

now equation of normal passing through point $P(\theta)$ is given by, 

$\dfrac{ax}{cos \theta} - \dfrac{by}{sin \theta} = a² - b². $

or, $\dfrac{\sqrt14x}{cos \theta} -\dfrac{ \sqrt5y}{sin \theta} = 14 - 5 = 9$ ....(1) 

it again meets the curve at the point $Q(2\theta) $

so, $Q(2\theta) = (√14cos2\theta, √5sin2\theta) $

now, put it in equation (1), 

or, $\dfrac{14cos2\theta}{\cos \theta} - \dfrac{5sin2\theta}{\sin\theta} = 9$ 

or, ${14(2cos² \theta - 1)}{\cos \theta} - \dfrac{10sin\theta cos \theta}{\sin \theta} = 9$

or, $28cos \theta - 14sec \theta - 10cos \theta = 9$

or, $18cos \theta - \dfrac{14}{cos \theta} = 9$

or, $18cos²\theta - 14 - 9cos \theta = 0$

or, $18cos²\theta -21cos \theta + 12cos\theta - 14 = 0$

or, $3cos \theta(6cos \theta - 7) + 2(cos \theta - 7) = 0$

or, $(3cos \theta + 2)(6cos \theta - 7) = 0$

or, $cos \theta = \dfrac{-2}{3} $


The number of tangents to the circle ${x}^{2}+{y}^{2}=3$ that are normals to the ellipse $\cfrac{{x}^{2}}{9}+\cfrac{{y}^{2}}{4}$ is

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

  1. one

  2. two

  3. three

  4. zero

Show answer
Correct Option: A
Explanation:
Let $y=mx+c$ is tangent to $x^2+y^2=3$ 
Then by condition of tangency
$\left|\dfrac{c}{\sqrt{m^2+1}}\right|=\sqrt{3}$
$\Rightarrow c^2=3(m^2+1)$        ...(i)
$y=mx+c$ is Normal to $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$
in $c=\dfrac{(b^2-a^2)m}{\sqrt{a^2+b^2m^2}}$

$\Rightarrow c=\dfrac{(4-9)m}{\sqrt{9+4m^2}}$

$\Rightarrow c^2=\dfrac{25m^2}{4m^2+9}$

$\Rightarrow 3(m^2+1)(4m^2+9)=25m^2$
let $m^2=t$
$\Rightarrow 3(t+1)(4t+9)=25t$
$\Rightarrow$ since, $D<0$.
Hence, There is no rout:
Hence, there is no tangent to circle which is Normal to ellipse.

Which of the following is/are true?

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

  1. There are infinite positive integral values of $a$ for which $(13x-1)^2+(13y-2)^2=\left (\dfrac {5x+112y-1}{a}\right )^2$ represents an ellipse

  2. The minimum distance of a point $(1, 2)$ from the ellipse $4x^2+9y^2+8x-36y+4=0$ is $1$

  3. If from a point $P(0, \alpha)$ two normals other than axes are drawn to the ellipse $\dfrac {x^2}{25}+\dfrac {y^2}{16}=1$, then $|\alpha| < \dfrac {9}{4}$

  4. If the length of latus rectum of an ellipse is one-third of its major axis, then its eccentricity is equal to $\dfrac {1}{\sqrt 3}$

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Correct Option: A,B,C

Number of distinct normal lines that can be drawn to the ellipse $\displaystyle \frac{x^2}{169} + \frac{y^2}{25} = 1$ from the point $P(0, 6)$ is:

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

  1. One

  2. Two

  3. Three

  4. Four

Show answer
Correct Option: C

If the normal at any point $P$ on the ellipse $\displaystyle\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ meets the axes in $G$ and $g$ respectively, then $PG:Pg=$

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

  1. $a:b$

  2. $a^2:b^2$

  3. $b:a$

  4. $b^2:a^2$

Show answer
Correct Option: D
Explanation:

Let $P\equiv (a\cos\theta, b\sin\theta)$
Thus equation of normal to the given ellipse at 'P' is given by,
$ax\sec\theta-by cosec\theta=a^2-b^2$
$\therefore G \equiv ((a-\cfrac{b^2}{a})\cos\theta,0), g\equiv (0,(b-\cfrac{a^2}{b})\sin\theta)$
Thus $PG = \sqrt{\cfrac{b^4}{a^2}\cos^2\theta+b^2\sin^2\theta}=\cfrac{b}{a}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
and $Pg = \sqrt{a^2\cos^2\theta+\cfrac{a^4}{b^2}\sin^2\theta}=\cfrac{a}{b}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
$\therefore PG:Pg = \cfrac{b^2}{a^2} $

The eccentricity of an ellipse whose centre is  at the origin is $\dfrac{1}{2}.$ If one of its directrices is $x =  - 4,$ then the equation of the normal to it at $\left( {1,\dfrac{3}{2}} \right)$ is

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

  1. $2y-x=2$

  2. $4x-2y=1$

  3. $4x+2y=7$

  4. $x+2y=4$

Show answer
Correct Option: B
Explanation:
Given: Eccentricity of ellipse$=\dfrac{1}{2}$
Now, $\dfrac{a}{e}=-4$
$\Rightarrow a=4\times\dfrac{1}{2}=2$
$\therefore {b}^{2}={a}^{2}\left(1-{e}^{2}\right)$
$\Rightarrow {a}^{2}\left(1-\dfrac{1}{4}\right)=3$
$\Rightarrow \dfrac{3}{4}{a}^{2}=3$
$\therefore {a}^{2}=4$
$\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}{3}=1$
Differentiating w.r.t $x$ we get
$\dfrac{2x}{4}+\dfrac{2y}{3}\dfrac{dy}{dx}=0$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{-2x}{4}}{\dfrac{2y}{3}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-3x}{4y}$
$\Rightarrow \left[\dfrac{dy}{dx}\right] _{\left(1,\frac{3}{2}\right)}=\dfrac{-3}{4}\times\dfrac{2}{3}=\dfrac{-1}{2}$
Equation of normal at $\left(1,\dfrac{3}{2}\right)$ is 
$y-\dfrac{3}{2}=2\left(x-1\right)$
$\Rightarrow 2y-3=4x-4$
$\Rightarrow 4x-2y=1$ is the equation of the normal.