Question
Let $P(10, 2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are S and $S'$. If the length of its latus rectum is 8, then the square of the area of $\Delta PSS'$ is equal to :
The correct answer is
900
Hyperbola Point and Latus Rectum Analysis
The problem requires finding the square of the area of a triangle $\Delta PSS'$, where P is a point on a hyperbola and S, S' are its foci. We are given the coordinates of P and the length of the latus rectum.
Calculating Hyperbola Parameters
The standard equation of the hyperbola is given as $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The point $P(10, 2\sqrt{15})$ lies on this hyperbola.
- Substitute Point P: Plugging the coordinates of P into the hyperbola equation yields: $ \frac{10^2}{a^2} - \frac{(2\sqrt{15})^2}{b^2} = 1 $ $ \frac{100}{a^2} - \frac{4 \times 15}{b^2} = 1 $ $ \frac{100}{a^2} - \frac{60}{b^2} = 1 \quad \quad (1) $
- Latus Rectum Length: The length of the latus rectum ($L$) is given as 8. For this hyperbola form, $L = \frac{2b^2}{a}$. $ \frac{2b^2}{a} = 8 $ $ b^2 = 4a \quad \quad (2) $
- Solve for $a$ and $b^2$: Substitute $b^2 = 4a$ from (2) into equation (1): $ \frac{100}{a^2} - \frac{60}{4a} = 1 $ $ \frac{100}{a^2} - \frac{15}{a} = 1 $ Multiplying the entire equation by $a^2$ gives: $ 100 - 15a = a^2 $ Rearranging this into a standard quadratic form: $ a^2 + 15a - 100 = 0 $ Factoring the quadratic equation: $ (a + 20)(a - 5) = 0 $ Since the semi-major axis $a$ must be positive, we have $a=5$. Now, find $b^2$ using equation (2): $ b^2 = 4a = 4(5) = 20 $
- Calculate $c^2$: For a hyperbola, the relationship between $a$, $b$, and $c$ (distance from center to focus) is $c^2 = a^2 + b^2$. $ c^2 = 5^2 + 20 = 25 + 20 = 45 $ Therefore, $c = \sqrt{45} = 3\sqrt{5}$.
Area of Triangle PSS' Calculation
The foci of the hyperbola are $S(-c, 0)$ and $S'(c, 0)$. The base of the triangle $\Delta PSS'$ is the distance between the foci, $SS' = 2c$. The height of the triangle is the perpendicular distance from point P to the line containing the foci (the x-axis), which is $|y_P|$.
Base length $SS' = 2c = 2(3\sqrt{5}) = 6\sqrt{5}$.
Height $= |y_P| = |2\sqrt{15}| = 2\sqrt{15}$.
The area of $\Delta PSS'$ is calculated using the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.
$ \text{Area} = \frac{1}{2} \times (2c) \times |y_P| = c \times |y_P| $ $ \text{Area} = (3\sqrt{5}) \times (2\sqrt{15}) $ $ \text{Area} = 6\sqrt{5 \times 15} = 6\sqrt{75} $ $ \text{Area} = 6\sqrt{25 \times 3} = 6 \times 5\sqrt{3} = 30\sqrt{3} $Square of the Area
The question asks for the square of the area of $\Delta PSS'$.
$ (\text{Area})^2 = (30\sqrt{3})^2 $ $ (\text{Area})^2 = 30^2 \times (\sqrt{3})^2 = 900 \times 3 = 2700 $Thus, the square of the area of $\Delta PSS'$ is 2700.
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