JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\theta\) be the acute angle between the tangents to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{1}=1\) and the circle \(x^{2}+y^{2}=3\) at their point of intersection in the first quadrant. Then \(\tan \theta\) is equal to :
- A \(\frac{5}{2 \sqrt{3}}\)
- B \(\frac{2}{\sqrt{3}}\)
- C \(\frac{4}{\sqrt{3}}\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
The point of intersection of the curves \(\frac{x^{2}}{9}+\frac{y^{2}}{1}=1\) and \(x^{2}+y^{2}=3\) in the first quadrant is \(\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)\) Now slope of tangent to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{1}=1\) at…
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