Suppose that the foci of the ellipse are and where and . Let and be two parabolas with a common vertex at (0, 0) and with foci at and , respectively. Let be a tangent to which passes through and be a tangent to which passes through . If is th

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 two 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 [2015]

Smart Achievers · Accepted Answer

(4)

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

$\Rightarrow a = 3, b = \sqrt{5}, e = \frac{2}{3} \Rightarrow f_{1} = 2, f_{2} = - 2$

$P_{1} : y^{2} = 8 x and P_{2} : y^{2} = - 16 x$

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

$It passes through (- 4, 0),$ $∴$ $0 = - 4 m_{1} + \frac{2}{m_{1}} \Rightarrow m_{1}^{2} = \frac{1}{2}$

$T_{2} : y = m_{2} x - \frac{4}{m_{2}}, It passes through (2, 0)$

$∴$ $0 = 2 m_{2} - \frac{4}{m_{2}} \Rightarrow m_{2}^{2} = 2$

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

Solution

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Solution

Ans. (4) Given: Ellipse is x29+y25=1 ⇒a=3, b=5, e=23⇒f1=2, f2=-2 P1: y2=8x and P2: y2=-16x T1: y=m1x+2m1 It passes through (-4,0), ∴ 0=-4m1+2m1⇒m12=12 T2: y=m2x-4m2, It passes through (2,0) ∴ 0=2m2-4m2⇒m22=2 ∴ 1m12+m22=2+2=4

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