Tag: normal to an ellipse

Questions Related to normal to an ellipse

Tangents are drawn to the ellipse $ \displaystyle \frac{x^2}{a^2}+\displaystyle \frac{y^2}{b^2}=1 $ at points where it is intersected by the line $ \ell x+my+n=0 $. Find the point of intersection of tangents at these points.

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

  1. $ \displaystyle \frac{-a^2}{n},\displaystyle \frac{-b^2m}{n} $

  2. $ \displaystyle \frac{-a^2\ell}{n},\displaystyle \frac{-b^2m}{n} $

  3. $ \displaystyle \frac{-a^2\ell}{n},\displaystyle \frac{-b^2}{n} $

  4. None of these

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Correct Option: B
Explanation:

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be the point of intersection of the
Line $lx+my+n=0$ and the ellipse $\cfrac { { x }^{ 2 }

}{ {a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Then the equation of tangent at $P$ is
$\cfrac

{ { xx } _{ 1 } }{ { a }^{ 2 } } +\cfrac { { yy } _{ 1 } }{ { b }^{ 2 } }

=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Since $\left( { x } _{ 1 },{ y } _{ 1 } \right) $  is the point of intersection of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Clearly $(i)$ and $(ii)$ represent the same line. Therefore,
$\therefore \quad \cfrac { { x } _{ 1 } }{ { a }^{ 2 }l } =\cfrac { { y } _{ 1 } }{ { b }^{ 2 }m } =\cfrac { 1 }{ -n } $
${ x } _{ 1 }=\cfrac { { -a }^{ 2 }l }{ n } ,\quad { y } _{ 1 }=\cfrac { -{ b }^{ 2 }m }{ n } $
Therefore, the point of intersection of given line and  the given ellipse is
$\left(- \cfrac { { a }^{ 2 }l }{ n } ,\cfrac { { b }^{ 2 }m }{ n }  \right) \quad $
Hence, option 'B' is correct.

A ray emanating from the point $(4, 0)$ is incident on the ellipse $9x^2\, +\, 25y^2\, =\, 225$ at the point $P$ with abscissa $3$. Find the equation of the reflected ray after first reflection.

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

  1. $12x + 5y = 48$

  2. $12x - 5y = 48$

  3. $-12x - 35y = 48$

  4. $-12x + 35y = 48$

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Correct Option: C,D
Explanation:

Solution:

Given equation of ellipse: $9x^2+25y^2=225$
$\cfrac{x^2}{25}+\cfrac{y^2}{9}=1$
Let $P(3,y _1)$
or, $9\times3^2+25y _1^2=225$
or, $25y _1^2=144$
or, $y _1=\pm\cfrac{12}{5}$
$\cfrac{x^2}{25}+\cfrac{y^2}{9}=1$
$\cfrac{x^2}{5^2}+\cfrac{y^2}{3^2}=1$
or, $e^2=1-\cfrac{b^2}{a^2}$

or, $e^2=1-\cfrac{9}{25}=\cfrac{16}{25}$
or, $e=\cfrac{4}{5}$
$ae=5\times\cfrac45=4$
It means the given point $(4,0)$ is focus of the ellipse.
We know that rays emanating from the one focus passes through other focus i.e $(-4,0)$
So, Equation of reflected ray $\Rightarrow y=\cfrac{\cfrac{12}{5}-0}{3+4}(x+4)$
or, $35y=12x+48$
or, $-12x+35y=48$
and another equation of reflected ray$\Rightarrow y=\cfrac{\cfrac{-12}{5}-0}{3+4}(x+4)$
or, $35y=-12x-48$
or, $-12x-35y=48$

The tangent and normal to the ellipse $x^2\, +\, 4y^2\, =\, 4$ at a point $P(\theta)$ on it meet the major axis in $Q$ and $R$ respectively. If $QR = 2$, the eccentric angle $\theta$ of $P$ is given by 

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

  1. $\cos \theta\, =\, \pm\, \dfrac23$

  2. $\sin \theta\, =\, \pm\, \dfrac23$

  3. $\tan \theta\, =\, \pm\, \dfrac23$

  4. $\cot \theta\, =\, \pm\, \dfrac23$

Show answer
Correct Option: A
Explanation:

The equation of tangent to ellipse is $bx cos\theta+ay sin\theta=ab$

It meets major axis at $x=\dfrac{a}{cos\theta}$
The equation of normal to ellipse is $ax sec \theta -by cosec \theta=a^{2}-b^{2}$
It meets the major axis at $x=\dfrac{a^{2}-b^{2}}{a sec\theta}$
Here $a=2$ and $b=1$
So, point $Q$ is $\left(\dfrac{2}{cos\theta},0 \right)$ and point $R$ is $\left(\dfrac{3}{2 sec\theta},0\right)$
The distance between them $QR= \dfrac{3cos\theta}{2}-\dfrac{2}{cos\theta}=2$
$\Rightarrow 3\cos ^{ 2 }{ \theta  } -4cos \theta-4=0$
$\Rightarrow cos \theta=-\dfrac{2}{3}$