JEE Advanced · Mathematics · 13. Parabola

Suppose that the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ are $(f_{1}, 0) a n d (f_{2}, 0)$ where $f_{1} > 0 a n d f_{2} < 0 .$ Let $P_{1} a n d P_{2}$ 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

A 4
B 16
C 8
D 2

Verified Solution

Answer & Solution

Correct Answer

(A) 4

Step-by-step Solution

Detailed explanation

f_{1} \equiv (2, 0)

f_{2} \equiv (- 2, 0)

Parabolas are

y^{2} = 8 x

y^{2} = - 16 x

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

0 = - 4 m + \frac{2}{m}

\Rightarrow m^{2} = \frac{1}{2}

T_{2} : y = m x - \frac{4}{m}

2 m - \frac{4}{m} = 0

\Rightarrow m^{2} = 2

Now,

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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