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Section: Mathematics

Question : 78 of 117

Marks: +1, -0

\frac{x^{2}}{27}+y^{2}=1

(3 \sqrt{3} \cos \theta, \sin \theta)

\theta \in \left(0, \frac{\pi}{2}\right)

\theta

Solution:

Given, equation of ellipse is

\frac{x^{2}}{27}+y^{2}=1

Now, equation of tangent to the ellipse at

P(3 \sqrt{3} \cos \theta, \sin \theta)

is given by

\frac{3 \sqrt{3} \cos \theta \cdot x}{27}+\sin \theta \cdot y=1

\Rightarrow \;\; \frac{x \cos \theta}{3 \sqrt{3}}+y \sin \theta=1

...(i)

X

-intercept of Eq. (i) is

x=3 \sqrt{3} \sec \theta=O A

(say)

Y

-intercept of Eq. (i) is

y=\csc \theta=O B

(say)

\therefore

Sum of intercepts

=3 \sqrt{3} \sec \theta+\csc \theta=f(\theta)

(say)

\Rightarrow f'(\theta)=3 \sqrt{3} \sec \theta \tan \theta-\csc \theta \cdot \cot \theta

Put

f'(\theta)=0

\Rightarrow \;\; \frac{3 \sqrt{3} \sin \theta}{\cos^{2} \theta}=\frac{\cos \theta}{\sin^{2} \theta} \Rightarrow \frac{\sin \theta}{\cos^{2} \theta} \cdot \frac{\sin^{2} \theta}{\cos \theta}=\frac{1}{3 \sqrt{3}}

\Rightarrow \;\; \tan^{3} \theta=\frac{1}{3 \sqrt{3}} \Rightarrow \theta=\frac{\pi}{6}

\because f'(\theta)

changes sign from negative to positive when moving from left to right.

\therefore f(\theta)

will attain minima at

\theta=\frac{\pi}{6}

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