Q96 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Mathematics
The ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b>a)\) and the parabola \(y^2=4 a x\) cut at right angles. If \(e\) is the eccentricity of the ellipse, then \(2 e^2=\)
- Correct answerA1
- B\(\frac{1}{2}\)
- C\(\frac{1}{8}\)
- D\(\frac{1}{3}\)
Solution
Given equation of parabola is \(y^2=4 a x\), then slope of tangent to be given parabola at point of intersection \(\left(a t^2, 2 a t\right)\) is \(m_1=\frac{1}{t^{\prime}}\), and slope of tangent to the given ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at point of intersection \(\left(a t^2, 2 a t\right)\) is
\(\begin{aligned}
m_2 & =-\frac{b^2}{a^2} \times\left(\frac{t}{2}\right) \\
\because \quad m_1 m_2 & =-1 \\
\Rightarrow\left(\frac{1}{t}\right) \times\left(-\frac{b^2}{a^2} \times \frac{t}{2}\right) & =-1 \Rightarrow \frac{b^2}{a^2}=2 \\
\because \quad e^2 & =1-\frac{a^2}{b^2}=1-\frac{1}{2} \quad(\because b > a) \\
\Rightarrow \quad 2 e^2 & =1
\end{aligned}\)
Hence, option (a) is correct.
Hence, the correct answer is option: 'A'.