AP EAMCET · Maths · Ellipse
If the normal at one end of a latus rectum of the ellipse \(\frac{x^2}{32}+\frac{y^2}{b^2}=1\) passes through one end of the minor axis, then \(\frac{e^4}{1-e^2}=\)
(Here \(e\) is the eccentricity of the ellipse)
- A \(\frac{1}{2}\)
- B \(1\)
- C \(\frac{2}{3}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
Equation of normal at \((x_1, y_1)\) to \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2\). Let the end of latus rectum be \((ae, \frac{b^2}{a})\). Normal equation:…
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