AP EAMCET · Maths · Ellipse
Find the condition for the line \(a x+b y+c=0\) to be a normal to an ellipse \(\frac{x^2}{4}+\frac{y^2}{36}=1\)
- A \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{144}{c^2}\)
- B \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{128}{c^2}\)
- C \(\frac{1}{a^2}+\frac{9}{b^2}=\frac{256}{c^2}\)
- D \(\frac{1}{a^2}+\frac{9}{b^2}=\frac{32}{c^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{a^2}+\frac{9}{b^2}=\frac{256}{c^2}\)
Step-by-step Solution
Detailed explanation
Let a point \(P(2 \cos \theta, 6 \sin \theta)\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{36}=1\), so equation of normal to the ellipse at point \(P\) is…
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