AP EAMCET · Maths · Ellipse
The points of intersection of the perpendicular tangents drawn to the ellipse \(4 x^2+9 y^2=36\) lie on the curve.
- A \(x^2+y^2=13\)
- B \(x^2-y^2=5\)
- C \(x+y=5\)
- D \(\frac{x^2}{9}+\frac{y^2}{4}=1\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2=13\)
Step-by-step Solution
Detailed explanation
We have, ellipse \[ \begin{aligned} 4 x^2+9 y^2 & =36 \\ \Rightarrow \quad \frac{x^2}{9}+\frac{y^2}{4} & =1 \quad \Rightarrow \quad \frac{x^2}{3^2}+\frac{y^2}{2^2}=1 \end{aligned} \] So, \(\quad a^2=9\) and \(b^2=4\). The points of intersection of the perpendicular tangents lie…
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