AP EAMCET · Maths · Ellipse
The product of the perpendicular distances drawn from the points \((3,0)\) and \((-3,0)\) to the tangent of the ellipse \(\frac{x^2}{36}+\frac{y^2}{27}=1\) at \(\left(3, \frac{9}{2}\right)\) is
- A 36
- B 27
- C 9
- D 63
Answer & Solution
Correct Answer
(B) 27
Step-by-step Solution
Detailed explanation
Let \(P(6 \cos \theta, 3 \sqrt{3} \sin \theta)\) be any point on the ellipse \(\frac{x^2}{36}+\frac{y^2}{27}=1\). The equation of the tangent at \(P(6 \cos \theta, 3 \sqrt{3} \sin \theta)\) is \(\frac{x}{6} \cos \theta+\frac{y}{3 \sqrt{3}} \sin \theta=1\).…
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