Hyperbola Latus Rectum Right Angle Problem

 

❓ Question

Consider the hyperbola

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

having one of its foci at

$$P(-3,\,0).$$

If the latus rectum through its other focus subtends a right angle at point $P$, and

$$a^2b^2=\alpha\sqrt{2}-\beta, \quad \alpha,\beta\in\mathbb{N},$$

then the value of

$$\alpha+\beta$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This problem is a classic JEE Advanced–type geometry–algebra mix involving:

✔ Properties of a hyperbola
✔ Geometry of the latus rectum
✔ Right-angle condition using vectors or slopes

We proceed step by step.

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🔹 Step 1 — Identify hyperbola parameters

Given:

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

For this hyperbola:

• Centre = $(0,0)$
  

• Foci = $(\pm c,0)$
  

• Where:

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

One focus is at $P(-3,0)$, so:

$$c=3 \Rightarrow a^2+b^2=9 \quad (1)$$

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🔹 Step 2 — Coordinates of the other focus and its latus rectum

The other focus is at:

$$Q(3,0)$$

The latus rectum through $Q$ is perpendicular to the transverse axis and has equation:

$$x=3$$

For a hyperbola, the endpoints of the latus rectum are:

$$\left(3,\ \pm\frac{b^2}{a}\right)$$

Let these points be:

$$L_1\left(3,\frac{b^2}{a}\right), \quad L_2\left(3,-\frac{b^2}{a}\right)$$

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🔹 Step 3 — Right-angle condition at focus $P$

Given:
The latus rectum subtends a right angle at $P(-3,0)$.

That means:

$$\angle L_1PL_2=90^\circ$$

Using vector geometry:

$$\vec{PL_1}\cdot\vec{PL_2}=0$$

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🔹 Step 4 — Apply dot product

Vectors:

$$\vec{PL_1}=(6,\ \tfrac{b^2}{a}), \quad \vec{PL_2}=(6,\ -\tfrac{b^2}{a})$$

Dot product:

$$(6)(6)+\left(\tfrac{b^2}{a}\right)\left(-\tfrac{b^2}{a}\right)=0$$
$$36-\frac{b^4}{a^2}=0 \Rightarrow \frac{b^4}{a^2}=36$$

So:

$$b^4=36a^2 \Rightarrow b^2=6a \quad (2)$$

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🔹 Step 5 — Solve using equations (1) and (2)

From (1):

$$a^2+b^2=9$$

Substitute $b^2=6a$:

$$a^2+6a=9$$
$$a^2+6a-9=0$$

Solve:

$$a=\frac{-6+\sqrt{36+36}}{2}=-3+3\sqrt{\mathrm{2}}$$

(negative root rejected)

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🔹 Step 6 — Compute $a^2b^2$

From $b^2=6a$:

$$a^2b^2=6a^3$$

Substitute $a=3(\sqrt{2}-1)$

$$a^3=27(\sqrt{2}-1{)}^3=27(5\sqrt{2}-7)$$

So:

$$a^2b^2=6\times27(5\sqrt{2}-7) =162(5\sqrt{2}-7)$$
$$=810\sqrt{2}-1134$$

Thus:

$$\alpha=810,\quad \beta=1134$$

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✅ Final Answer

$$\alpha+\beta = 810+1134 = \boxed{1944}$$

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