Let θ be the acute angle between the tangents to the ellipse {x²}{9}+{y²}{1}=1 and the circle x²+y²=3 at their point of intersection in the first quadrant. Then, θ is equal to [1 Sep 2021 Shift 2]

Correct Answer is : 23{2}{{3}}32

Correct Answer is : 23{2}{{3}}32

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Section: Mathematics

Question : 66 of 117

Marks: +1, -0

\frac{x^{2}}{9}+\frac{y^{2}}{1}=1

x^{2}+y^{2}=3

\tan \theta

Solution:

Given, ellipse

\frac{x^{2}}{9}+\frac{y^{2}}{1}=1

...(i)

and circle

x^{2}+y^{2}=3

...(ii)

The point of intersection by solving Eqs. (i) and (ii) in

first quadrant

\left(\frac{3}{2},\frac{\sqrt{3}}{2}\right)

Differentiating Eqs. (i) and (ii) w.r.t. x, we have

Let

m_{1}=\frac{dy}{dx}=\frac{-x}{9y}

and

m_{2}=\frac{dy}{dx}=\frac{-x}{y}

\left(\frac{3}{2},\frac{\sqrt{3}}{2}\right)

m_{1}=-\frac{1}{3\sqrt{3}}, m_{2}=-\sqrt{3}

If angle between both curves is θ, then

\tan \theta =\left|\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right|

=\left|\frac{-\frac{1}{3\sqrt{3}}+\sqrt{3}}{1+\left(-\frac{1}{3\sqrt{3}}\right)(-\sqrt{3})}\right|=\frac{2}{\sqrt{3}}

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