Question

Consider a hyperbola H: x² - 2y² = 4. Let the tangent at a point P(4, √6) meet the x-axis at Q and latus rectum at R(x₁, y₁), x₁>0. If F is a focus of H which is nearer to the point P, then the area of ΔQFR is equal to :

• A. √6 - 1
• B. 7/√6 - 2
• C. 4√6 - 1
• D. 4√6

Hint:

For a hyperbola $x^2/a^2 - y^2/b^2 = 1$, the latus rectum equation is $x = \pm ae$.

Answer 1

The Correct Option is B

Step 1: Hyperbola: $\frac{x^2}{4} - \frac{y^2}{2} = 1$. $a^2=4, b^2=2$. $e = \sqrt{1 + b^2/a^2} = \sqrt{1 + 2/4} = \sqrt{3/2}$. Focus $F(ae, 0) = (2\sqrt{3/2}, 0) = (\sqrt{6}, 0)$.
Step 2: Tangent at $P(4, \sqrt{6})$: $\frac{4x}{4} - \frac{\sqrt{6}y}{2} = 1 \implies x - \frac{\sqrt{6}}{2}y = 1$. Meeting $x$-axis at $Q$: $y=0 \implies x=1 \implies Q(1, 0)$.
Step 3: Latus rectum is $x = \sqrt{6}$. Point $R$: $x = \sqrt{6} \implies \sqrt{6} - \frac{\sqrt{6}}{2}y = 1 \implies \frac{\sqrt{6}}{2}y = \sqrt{6}-1 \implies y = 2 - \frac{2}{\sqrt{6}}$. $R(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$.
Step 4: $\Delta QFR$ vertices: $Q(1,0), F(\sqrt{6}, 0), R(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$. Area $= \frac{1}{2} \cdot \text{Base} \cdot \text{Height} = \frac{1}{2} \cdot (\sqrt{6} - 1) \cdot (2 - \frac{2}{\sqrt{6}}) = (\sqrt{6}-1)(1 - \frac{1}{\sqrt{6}}) = \sqrt{6} - 1 - 1 + \frac{1}{\sqrt{6}} = \sqrt{6} + \frac{1}{\sqrt{6}} - 2 = \frac{7}{\sqrt{6}} - 2$.