TS EAMCET · Maths · Ellipse
If the product of the lengths of the perpendiculars drawn from the foci to the tangent \(y=\frac{-3}{4} x+3 \sqrt{2}\) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is 9 , then the eccentricity of that ellipse is
- A \(\frac{\sqrt{2}}{3}\)
- B \(\frac{\sqrt{5}}{6}\)
- C \(\frac{1}{9}\)
- D \(\frac{\sqrt{7}}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{7}}{4}\)
Step-by-step Solution
Detailed explanation
We know that, the product of the length of the perpendicular drawn from the foci to the tangent of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is \(b^2\). Also, length \(=9 \quad \therefore \quad b^2=9\) Equation of tangent \(y=\frac{-3}{4} x+3 \sqrt{2}\)…
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