TS EAMCET · Maths · Ellipse
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b>a)\) is an ellipse with eccentricity \(\frac{1}{\sqrt{2}}\). If the angle of intersection between the ellipse and parabola \(y^2=4 a x\) is \(\theta\), then the coordinates of the point \(\frac{2 \theta}{3}\) on the ellipse is
- A \(\left(\frac{a}{2}, \frac{a}{2}\right)\)
- B \(\left(\frac{a}{2}, \frac{3 a}{2}\right)\)
- C \(\left(\frac{\sqrt{3} a}{2}, \frac{3 \sqrt{3} a}{\sqrt{2}}\right)\)
- D \(\left(\frac{a}{2}, \frac{\sqrt{3} a}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{a}{2}, \frac{\sqrt{3} a}{\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
\(e^2 = 1 - \frac{a^2}{b^2}\) \(\frac{1}{2} = 1 - \frac{a^2}{b^2} \Rightarrow b^2 = 2a^2\) Slope of tangent to ellipse: \(m_1 = -\frac{b^2 x_0}{a^2 y_0}\) Slope of tangent to parabola: \(m_2 = \frac{2a}{y_0}\) For orthogonality, \(m_1 m_2 = -1\)…
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