TS EAMCET · Maths · Ellipse
If the perpendicular distance from the focus of an ellipse \(\frac{x^2}{9}+\frac{y^2}{b^2}=1(b < 3)\) to its corresponding directrix is \(\frac{4}{\sqrt{5}}\), then the slope of the tangent to this ellipse drawn at \(\left(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\) is
- A \(-\frac{2}{3}\)
- B \(\frac{2}{3}\)
- C \(\frac{3}{2}\)
- D \(-\frac{3}{2}\)
Answer & Solution
Correct Answer
(A) \(-\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(a=3\) \(\frac{a(1-e^2)}{e} = \frac{4}{\sqrt{5}}\) \(b^2=a^2(1-e^2) \Rightarrow 1-e^2=\frac{b^2}{9}\) \(\frac{3(b^2/9)}{e} = \frac{4}{\sqrt{5}} \Rightarrow \frac{b^2}{3e} = \frac{4}{\sqrt{5}} \Rightarrow b^2 = \frac{12e}{\sqrt{5}}\)…
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