First slide

Tangents and normals

Question

$The coordinates of the point P (x, y) lying in the first quadrant on the$ $ellipse x^{2} / 8 + y^{2} / 18 = 1 so that the area of the triangle formed by the$ $tangent at P and the coordinate axes is the smallest, are given by$

Moderate

A

(2, 3)
B

$(\sqrt{8}, 0)$
C

$(\sqrt{18}, 0)$
D

None of these

Solution

$\begin{array}{l} Any point on the ellipse is given by (\sqrt{8} \cos θ, \sqrt{18} \sin θ) \\ Now \frac{2 x}{8} + \frac{2}{18} y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 9 x}{4 y} \\ {\Rightarrow \frac{d y}{d x}|}_{(\sqrt{8} \cos θ, \sqrt{18} \sin θ)} = - \frac{9 \sqrt{8} \cos θ}{4 \sqrt{18} \sin θ} \end{array}$
$= - \frac{\sqrt{9}}{2} \cot θ$
$Hence the equation of the tangent at (\sqrt{8} \cos θ, \sqrt{18} \sin θ) is$
$y - \sqrt{18} \sin θ = \frac{\sqrt{9}}{2} \cot θ (x - \sqrt{8} \cos θ)$
Therefore, the tangent cuts the coordinate axes at the points
$(0, \frac{\sqrt{18}}{\sin θ}) a n d (\frac{\sqrt{8}}{\cos θ}, 0)$
Thus the area of the triangle formed by this tangent and the coordinate axes is
$\begin{aligned} A = \frac{1}{2} \sqrt{18} \cdot \sqrt{8} \cdot \frac{1}{\cos θ \sin θ} \\ = \frac{6}{\cos θ \sin θ} = 12 cosec 2 θ \end{aligned}$
$But cosec 2 θ is smallest when θ = π / 4 . Therefore A is$
$smallest when θ = π / 4$
Hence the required point is $(\sqrt{8} \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{18} \cdot \frac{1}{\sqrt{2}}) = (2, 3)$

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