Question

If the eccentricity of the hyperbola x² – y²sec²θ = 4 is $\sqrt{3}$ times the eccentricity of the ellipse x²sec²θ + y² = 16 the value of θ equals

• A. π/6
• B. 3π/4
• C. π/3
• D. π/2

Answer 1

Answer: (B)

3π/4

Answer 2

Given x² – y²sec²θ = 4

⇒ $\frac{x^{2}}{4} - \frac{y^{2}}{4\cos^{2}\theta} = 1\mspace{6mu} and\mspace{6mu}\frac{x^{2}}{16\cos^{2}\theta} + \frac{y^{2}}{16} = 1$According to problem $\frac{4 + 4\cos^{2}\theta}{4} = 3\left( \frac{16 - 16\cos^{2}\theta}{16} \right)$

⇒ 1 + cos²θ = 3(1 – cos²θ)

⇒ 4cos²θ = 2

cosθ = $\pm \frac{1}{\sqrt{2}}$

θ = π/4, $\frac{3\pi}{4}$