Question

The area of the triangle formed by the tangent at any point of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the axes is minimum at the point

• A. $(\frac{a}{\sqrt{2}}\frac{b}{\sqrt{2}})$
• B. $(\frac{a}{2},\frac{b}{2})$
• C. $(a,b)$
• D. $(\frac{\sqrt{a}}{2}\frac{\sqrt{b}}{2})$

Answer 1

Answer: (A)

$(\frac{a}{\sqrt{2}}\frac{b}{\sqrt{2}})$

Let any point $(asin\theta ,bcos\theta )$

$\frac{2x}{a^2}+\frac{2y{y}^1}{b^2}=0$

$y^1=\frac{-x{b}^2}{a^{2}y}$

$y^1=\frac{-btan\theta }{a}$

$(y-bcos\theta )=\frac{-btan\theta }{a}(x-asin\theta )$

$\frac{aco{s}^2\theta }{sin\theta }+asin\theta =x$ (x intercept)

$y=bcos\theta +\frac{bsi{n}^2\theta }{cos\theta }$ (y intercept)

$area=\frac{1}{2}xy$

$=\frac{1ab}{2sin\theta cos\theta }$

$=\frac{ab}{si{n}^2\theta }$

area is minimum when $si{n}^2\theta =1$

So $\theta =\frac{\pi }{4}$