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Suppose that the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ are $\left(f_1, 0\right)$ and $\left(f_2, 0\right)$ 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 $\left(f_1, 0\right)$ and $\left(2 f_2, 0\right)$, respectively. Let $T_1$ be a tangent to $P_1$ which passes through $\left(2 f_2, 0\right)$ and $T_2$ be a tangent to $P_2$ which passes through $\left(f_1, 0\right)$. The $m_1$ is the slope of $T_1$ and $m_2$ is the slope of $T_2$, then the value of $\left(\frac{1}{m^2}+m_2^2\right)$ is
$1$
$2$
$3$
$4$
Solution
The equation of $P_1$ is $y^2-8 x=0$ and $P_2$ is $y^2+16 x=0$
Tangent to $y ^2-8 x =0$ passes through $(-4,0)$
$\Rightarrow 0= m _1(-4)+\frac{2}{ m _1} \Rightarrow \frac{1}{ m _1^2}=2$
Also tangent to $y^2+16 x=0$ passes through $(2,0)$
$\Rightarrow 0= m _2 \times 2-\frac{4}{ m _2} \Rightarrow m _2^2=2$
$\Rightarrow \frac{1}{ m _1^2}+ m _2^2=4$
Similar Questions
Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
| $List-I$ | $List-II$ |
| If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
| If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
| If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
| If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
| ($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is:
