Tag: normal to an ellipse

Questions Related to normal to an ellipse

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

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

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

  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

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Correct Option: B
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,
$\cfrac{16x}{2\sqrt{3}}-\cfrac{4y}{1}=12$
$\Rightarrow 2x=\sqrt{3}(y+3)$

If the tangent drawn at a point $\left( { t }^{ 2 },2t \right) $ on the parabola ${ y }^{ 2 }=4x$ is same as normal drawn at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ on the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$, then which of following is true.

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

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

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

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

  4. None of these

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Correct Option: A,B,C
Explanation:

Equation of tangent to ${ y }^{ 2 }=4x$ at $\left( { t }^{ 2 },2t \right) $ is $x=ty-{ t }^{ 2 }$   ....(1)


Equation of normal to ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$ at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ is $\sqrt { 5 } \sec { \alpha x-2y\csc { \alpha  } =1 } $    ....(2)

Given (1) $=$ (2)

$\displaystyle \Rightarrow \sqrt { 5 } \sec { \alpha  } =\frac { 2\csc { \alpha  }  }{ t } =-\frac { 1 }{ { t }^{ 2 } } \Rightarrow \cos { \alpha  } =-\sqrt { 5 } { t }^{ 2 }$ and $\sin { \alpha  } =-2t$

$\displaystyle \Rightarrow \cos ^{ 2 }{ \alpha  } +\sin ^{ 2 }{ \alpha  } =5{ t }^{ 4 }+4{ t }^{ 2 }=1\Rightarrow { t }^{ 2 }=\frac { 1 }{ 5 } $   

$\therefore$ (A) is true
and $\displaystyle \frac { \sin { \alpha  }  }{ \cos { \alpha  }  } =-\frac { 2t }{ -\sqrt { 5 } { t }^{ 2 } } =\frac { 2 }{ \sqrt { 5 }  } \times \frac { 1 }{ t } =\frac { 2 }{ \sqrt { 5 }  } \times \left( \pm 5 \right) $

$\therefore \tan { \alpha  } =\pm 2$

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

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

  1. $3$

  2. $4$

  3. $2$

  4. $1$

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Correct Option: D
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 $1$.

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

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

  1. $3 x - y = \sqrt { 5 }$

  2. $3 x - y = \sqrt { 5 } - 3$

  3. $3 x - y = \sqrt { 5 } + 2$

  4. $3 x - y = \sqrt { 5 } ( \sqrt { 5 } + 1 )$

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Correct Option: A

The number of normals that can be drawn to the curve $\displaystyle 4x^{2} + 9y^{2} = 36$ from an external point, in general, is

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

  1. $1$

  2. $3$

  3. $4$

  4. infinite

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Correct Option: C
Explanation:

Given ellipse may be written as, $\displaystyle \cfrac{x^{2}}{9} + \cfrac{y^{2}}{4} = 1$
$\Rightarrow a^2=9, b^2=4$
Thus general equation of normal to ellipse with slope $m$ is given by,
$y = mx-\cfrac{(a^2-b^2)m}{\sqrt{a^2+b^2m^2}}=mx-\cfrac{5m}{\sqrt{9+4m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{25m^2}{9+4m^2}$
$\Rightarrow (y _1-mx _1)^2(9+4m^2)=25m^2$
Clearly this is a polynomial of degree four so maximum number of normal that can be drawn from point $P$ (any external point) to the ellipse is $4$ corresponding to four roots of $m.$

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 $p+q+r.$

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

  1. $8$

  2. $9$

  3. $10$

  4. $12$

Show answer
Correct Option: C
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, $p+q+r=10$

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

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

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

Show answer
Correct Option: C
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.