Question

A straight line

P Q

touches the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and the circle

x^{2} + y^{2} = r^{2} (b < r < a)

.

R S

is a focal chord of the ellipse. If

R S

is parallel to

P Q

and meets the circle at points

R

and

S

. Find the length of

R S

.

Accepted Answer

Equation of tangent to the ellipse in slope form is $y = m x \pm \sqrt{a^{2} m^{2} + b^{2}}$

This is also tangent to the circle with centre $(0, 0)$ .

radius $= r = \pm \frac{\sqrt{a^{2} m^{2} + b^{2}}}{\sqrt{1 + m^{2}}}$ (Using Distance formula of a line from a point)

$\Rightarrow r^{2} = \frac{a^{2} m^{2} + b^{2}}{1 + m^{2}} \dots (1)$

$\Rightarrow r^{2} + r^{2} m^{2} = a^{2} m^{2} + b^{2}$

$\Rightarrow m^{2} (r^{2} - a^{2}) = b^{2} - r^{2}$

$\Rightarrow m = \sqrt{\frac{b^{2} - r^{2}}{(r^{2} - a^{2})}}$

Chord $R S$ passes through focus of ellipse $(a e, 0)$ and having slope equal to the slope of $P Q$ .

Equation of $R S$ is $y - 0 = m (x - a e)$

Draw a line from $O$ to the mid-point of $R S$ .

Therefore, $O T = \frac{|a m e|}{\sqrt{1 + m^{2}}}$ (Using Distance formula of a line from a point)

By Pythagoras theorem, As $∆ R O T$ is right-angled at $T$

$\Rightarrow O R^{2} = O T^{2} + R T^{2}$

$\Rightarrow R T^{2} = O T^{2} - O R^{2}$

$\Rightarrow R T^{2} = r^{2} - {(\frac{|a m e|}{\sqrt{1 + m^{2}}})}^{2}$

$= \frac{r^{2} (1 + m^{2}) - a^{2} m^{2} e^{2}}{1 + m^{2}}$

$= \frac{(a^{2} m^{2} + b^{2}) - a^{2} m^{2} e^{2}}{1 + m^{2}}$ (From Equation $1$ )

$= \frac{a^{2} m^{2} + b^{2} - m^{2} (a^{2} - b^{2})}{1 + m^{2}}$ $(b^{2} = a^{2} - a^{2} e^{2})$

$= \frac{b^{2} (1 + m^{2})}{1 + m^{2}} = b^{2}$

$\Rightarrow R T = b$

$R S = 2 R T = 2 b$

Embibe Experts Solutions for Chapter: Ellipse, Exercise 4: Exercise-4

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Ellipse, Exercise 4: Exercise-4

Questions from Embibe Experts Solutions for Chapter: Ellipse, Exercise 4: Exercise-4 with Hints & Solutions

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