TS EAMCET · Maths · Ellipse
\(L_1^{\prime}\) is the end of a latus rectum of the ellipse \(3 x^2+4 y^2\) \(=12\) which is lying in the third quadrant. If the normal drawn at \(L_1^{\prime}\) to this ellipse intersects the ellipse again at the point \(P(a, b)\) then \(a=\)
- A \(\frac{63}{38}\)
- B \(\frac{11}{19}\)
- C \(-\frac{11}{19}\)
- D \(-\frac{63}{38}\)
Answer & Solution
Correct Answer
(B) \(\frac{11}{19}\)
Step-by-step Solution
Detailed explanation
\(3 x^2+4 y^2=12 \qquad ....\mathrm{(i)}\) \(\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow a^2=4, b^2=3, e=\frac{1}{2}\) End of latus rectum in third quadrant is \(\left(-a e,-\frac{b^2}{a}\right)\) \(\Rightarrow L_1^{\prime}=\left(-1, \frac{-3}{2}\right)\) Eq. of normal…
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