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Standard XII

Mathematics

Focii of Ellipse

Suppose that ...

Question

Suppose that the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ are $(f_{1}, 0)$ and $(f_{2}, 0)$ where $f_{1} > 0$ and $f_{2} < 0$ . Let $P_{1}$ and $P_{2}$ be the parabolas with a common vertex at $(0, 0)$ and with foci at $(f_{1}, 0)$ and $(2 f_{2}, 0)$ respectively. Let $T_{1}$ be a tangent to $P_{1}$ which passes through $(2 f_{2}, 0)$ and $T_{2}$ be a tangent to $P_{2}$ which passes through $(f_{1}, 0)$ . If $m_{1}$ is the slope of $T_{1}$ and $m_{2}$ is the slope of $T_{2}$ , then the value of $(\frac{1}{m_{1}^{2}} + m_{2}^{2})$ is

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Solution

Given ellipse is $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$

$e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{5}{9} = \frac{4}{9} \Rightarrow e = \frac{2}{3}$

Foci of the ellipse are $(\pm a e, 0)$

So, $(f_{1}, 0)$ is $(2, 0)$ and $(f_{2}, 0)$ is $(- 2, 0)$

Foci of parabola $P_{1}$ is $(2, 0)$ and foci of $P_{2}$ is $(- 4, 0)$ .

Equation of tangent to any parabola in slope form is $y = m x + \frac{a}{m}$ .

Therefore, the equation of tangent to $P_{1}$ is

$y = m_{1} x + \frac{2}{m_{1}}$

It passes through $(- 4, 0)$ .

$\Rightarrow 0 = - 4 m_{1} + \frac{2}{m_{1}} \Rightarrow m_{1}^{2} = \frac{1}{2} \Rightarrow \frac{1}{m_{1}^{2}} = 2$ .......(i)

Equation of tangent to $P_{2}$ is

$y = m_{2} x + \frac{- 4}{m_{2}} \Rightarrow 0 = 2 m_{2} + \frac{- 4}{m_{2}} \Rightarrow m_{2}^{2} = 2$ ......(ii)

Adding (i) and (ii), we get

$\frac{1}{m_{1}^{2}} + m_{2}^{2} = 2 + 2 \frac{1}{m_{1}^{2}} + m_{2}^{2} = 4$

So, correct answer is $4$ .

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\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1

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f_{1} > 0

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