Find Product of Focal Distances Hyperbola

 ❓ Question

Let a hyperbola in standard form have transverse and conjugate axes of lengths $2a$ and $2b$, respectively. One focus is at $(-5,0)$ and the corresponding directrix is $5x + 9 = 0$.

If the product of the focal distances of the point $(\alpha ,2\sqrt{5})\; on\; the\; hyperbola\; is$$p$, find $4p$.

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Hyperbola standard form
👉 Focus-directrix property
👉 Product of focal distances.

Main idea:

For hyperbola:

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$
$$e=\frac ca$$

and directrix:

$$x=\pm\frac ae$$

Also:

$$c^2=a^2+b^2$$

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🔷 Step 1 — Identify Focus and Directrix 💯

Given focus:

$$(-5,0)$$

Thus:

$$c=5$$

Given directrix:

$$5x+9=0$$
$$x=-\frac95$$

For hyperbola:

$$x=-\frac ae$$

Hence:

$$\frac ae=\frac95$$

But:

$$e=\frac ca=\frac5a$$

Substitute:

$$\frac a{5/a}=\frac95$$
$$\frac{a^2}{5}=\frac95$$
$$a^2=9$$
$$a=3$$

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🔷 Step 2 — Find $b^2$

Using:

$$c^2=a^2+b^2$$
$$25=9+b^2$$
$$b^2=16$$

Thus hyperbola is:

$$\boxed{ \frac{x^2}{9}-\frac{y^2}{16}=1 }$$

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🔷 Step 3 — Find Point on Hyperbola

Given point:

$$(\alpha,2\sqrt5)$$

Substitute into hyperbola:

$$\frac{\alpha^2}{9} - \frac{(2\sqrt5)^2}{16} =1$$
$$\frac{\alpha^2}{9}-\frac{20}{16}=1$$
$$\frac{\alpha^2}{9}-\frac54=1$$
$$\frac{\alpha^2}{9}=\frac94$$
$$\alpha^2=\frac{81}{4}$$
$$\alpha=\pm\frac92$$

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🔷 Step 4 — Product of Focal Distances

For hyperbola:

$$\boxed{ S_1S_2=b^2 }$$

where:

$$S_1,S_2$$

are focal distances of any point on hyperbola.

Thus:

$$p=b^2=16$$

Hence:

$$4p=4\times16$$
$$=64$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Directrix formula ellipse wala use kar lena

❌ $c^2=a^2-b^2$ use kar dena instead of hyperbola relation

❌ Product of focal distances property bhool jaana

Remember:

For hyperbola:

$$\boxed{ c^2=a^2+b^2 }$$

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✅ Final Answer

$$\boxed{64}$$

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