If the eccentricity of the hyperbola $x^{2} - y^{2}\sec^{2}{ heta = 4}\ is\ \sqrt{3}$ times the eccentricity of the ellipse $x^{2}\sec^{2}{ heta + y^{2} = 16,}$ then the value of $ heta$ equals

Question

If the eccentricity of the hyperbola $x^{2} - y^{2}\sec^{2}{	heta = 4}\ is\ \sqrt{3}$ times the eccentricity of the ellipse $x^{2}\sec^{2}{	heta + y^{2} = 16,}$ then the value of $	heta$ equals

Dr. Abubacker · Accepted Answer

The correct answer is Option B: $\frac{3\pi}{4}$.

Q. If the eccentricity of the hyperbola $x^{2} - y^{2}\sec^{2}{\theta = 4}\ is\ \sqrt{3}$ times the eccentricity of the ellipse $x^{2}\sec^{2}{\theta + y^{2} = 16,}$ then the value of $\theta$ equals

Explanation

Reviewed by Dr. Abubacker

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