TS EAMCET · Maths · Ellipse
A line perpendicular to the \(X\)-axis cuts the circle \(x^2+y^2=9\) at \(A\) and the ellipse \(4 x^2+9 y^2=36\) at \(B\) such that \(A\) and \(B\) lie in the same quadrant. If \(\theta\) is the greatest acute angle between the tangents drawn to the curves at \(A\) and \(B\), then \(\tan \theta=\)
- A \(\frac{1}{12}\)
- B \(\frac{1}{2 \sqrt{6}}\)
- C \(\frac{5}{24}\)
- D \(\frac{5}{4 \sqrt{6}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2 \sqrt{6}}\)
Step-by-step Solution
Detailed explanation
Let the equation of line perpendicular to \(X\)-axis cut the circle \(x^2+y^2=9\) at \(A\) is \((3 \cos \alpha, 3 \sin \alpha)\) Equation of tangent at \(A\) is \(x \cos \alpha+y \sin \alpha=3\) Slope \(=-\cot \alpha\) Similarly, cut the ellipse \(4 x^2+9 y^2=36\) at \(B\) is…
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