Let a tangent be drawn to the ellipse [x^2 27 + y^2 = 1] at [(3 3 , )] where [0 ( 0, 2 )].

Question

Let a tangent be drawn to the ellipse $${{{x^2}} \over {27}} + {y^2} = 1$$ at $$(3\sqrt 3 \cos \theta ,\sin \theta )$$ where $$0 \in \left( {0,{\pi  \over 2}} \right)$$. Then the value of $$\theta$$ such that the sum of intercepts on axes made by this tangent is minimum is equal to :

Accepted Answer

Tangent = $${x \over {3\sqrt 3 }}\cos \theta  + y\sin \theta  = 1$$

x-intercept = $${3\sqrt 3 }$$ sec$$\theta$$

y-intercept = cosec$$\theta$$

sum = $${3\sqrt 3 }$$ sec$$\theta$$ + cosec$$\theta$$ = f($$\theta$$) $$\theta$$ $\in$$ $\left( {0,{\pi  \over 2}} \right)$$

$$ \Rightarrow $$ f'($$\theta$$) = $${3\sqrt 3 }$$ sec$$\theta$$tan$$\theta$$ - cosec$$\theta$$ cot$$\theta$$ = 0

$$ \Rightarrow $$ $${{3\sqrt 3 \sin \theta } \over {{{\cos }^2}\theta }} = {{\cos \theta } \over {\sin \theta }}$$

$$ \Rightarrow {\tan ^3}\theta  = {\left( {{1 \over {\sqrt 3 }}} \right)^3}$$

$$ \Rightarrow \tan \theta  = {1 \over {\sqrt 3 }}$$

$$ \Rightarrow $$ $$\theta$$ = $${{\pi  \over 6}}$$

also f'($$\theta$$) changes sign - to + hence minimum.

Solutions · 1