Question

The eccentricity of the ellipse \[ p x^2 + 5y^2 = 80, \quad \text{where } p > 5, \] is \( \frac{\sqrt{3}}{2} \). Then the value of \( p \) is

• A. \( \frac{5}{8} \)
• B. \( 16 \)
• C. \( \frac{5}{4} \)
• D. \( 20 \)
• E. \( 25 \)

Hint:

For an ellipse, use \( e = \frac{\sqrt{a^2 - b^2}}{a} \) to find eccentricity.

Answer 1

The Correct Option is D

The standard form of an ellipse equation is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] Given: \[ p x^2 + 5 y^2 = 80 \] Dividing by 80: \[ \frac{x^2}{\frac{80}{p}} + \frac{y^2}{\frac{80}{5}} = 1 \] \[ \frac{x^2}{\frac{80}{p}} + \frac{y^2}{16} = 1 \] Comparing with the standard form: \[ a^2 = \frac{80}{p}, \quad b^2 = 16 \] Eccentricity of an ellipse: \[ e = \frac{\sqrt{a^2 - b^2}}{a} \] Given \( e = \frac{\sqrt{3}}{2} \), so: \[ \frac{\sqrt{\frac{80}{p} - 16}}{\sqrt{\frac{80}{p}}} = \frac{\sqrt{3}}{2} \] Squaring both sides: \[ \frac{\frac{80}{p} - 16}{\frac{80}{p}} = \frac{3}{4} \] \[ 1 - \frac{16p}{80} = \frac{3}{4} \] \[ 1 - \frac{p}{5} = \frac{3}{4} \] \[ \frac{4}{4} - \frac{p}{5} = \frac{3}{4} \] \[ p = 20 \] 
Final Answer: \[ \boxed{20} \]