Coordinate Geometry – Hyperbola – JEE Main 22 Jan 2026 Shift 2

Solution:

The equation of the hyperbola is $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$.

Given the length of the latus rectum is 8:
$$ \frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a $$

The point $P(10, 2\sqrt{15})$ lies on the hyperbola, so it must satisfy the equation:
$$ \frac{100}{a^2} – \frac{(2\sqrt{15})^2}{b^2} = 1 $$
$$ \frac{100}{a^2} – \frac{60}{b^2} = 1 $$

Substitute $b^2 = 4a$ into the equation:
$$ \frac{100}{a^2} – \frac{60}{4a} = 1 $$
$$ \frac{100}{a^2} – \frac{15}{a} = 1 $$
Multiplying by $a^2$:
$$ 100 – 15a = a^2 $$
$$ a^2 + 15a – 100 = 0 $$
$$ (a + 20)(a – 5) = 0 $$
Since $a > 0$, we have $a = 5$.

Now find $b^2$:
$$ b^2 = 4(5) = 20 $$

To find the area of $\Delta PSS’$, we need the coordinates of the foci $S$ and $S’$.
The eccentricity $e$ is given by $b^2 = a^2(e^2 – 1)$:
$$ 20 = 25(e^2 – 1) \Rightarrow \frac{4}{5} = e^2 – 1 \Rightarrow e^2 = \frac{9}{5} \Rightarrow e = \frac{3}{\sqrt{5}} $$
The foci are $(\pm ae, 0)$:
$$ ae = 5 \times \frac{3}{\sqrt{5}} = 3\sqrt{5} $$
So, $S(3\sqrt{5}, 0)$ and $S'(-3\sqrt{5}, 0)$. The base of the triangle $SS’ = 6\sqrt{5}$.

The height of the triangle is the y-coordinate of P, which is $2\sqrt{15}$.
$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$
$$ \text{Area} = \frac{1}{2} \times (6\sqrt{5}) \times (2\sqrt{15}) $$
$$ \text{Area} = 6\sqrt{75} = 6 \times 5\sqrt{3} = 30\sqrt{3} $$

The square of the area is:
$$ (30\sqrt{3})^2 = 900 \times 3 = 2700 $$

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