Solveeit Logo
Login

Question

Question: The coordinates of the point \(P(x,y)\) lying in the first quadrant on the ellipse \(\frac{x^{2}}{8}...

The coordinates of the point P(x,y)P(x,y) lying in the first quadrant on the ellipse x28+y218=1\frac{x^{2}}{8} + \frac{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

A

(2,3)

B

(8,0)\left( \sqrt{8},0 \right)

C

(18,0)\left( \sqrt{18},0 \right)

D

None

Answer

(2,3)

Explanation

Solution

Any point on the ellipse is given by

(8cos⁡θ,18sin⁡θ)\left( \sqrt{8}\cos\theta,\sqrt{18}\sin\theta \right)

Now 2x8+218ydydx=0\frac{2x}{8} + \frac{2}{18}y\frac{dy}{dx} = 0 ⇒dydx=−9x4y\frac{dy}{dx} = - \frac{9x}{4y}

⇒  dydx∣(8cos⁡θ,18sin⁡θ)=−98cos⁡θ418sin⁡θ=−92cot⁡θ\left. \ \frac{dy}{dx} \right|_{\left( \sqrt{8}\cos\theta,\sqrt{18}\sin\theta \right)} = - \frac{9\sqrt{8}\cos\theta}{4\sqrt{18}\sin\theta} = - \frac{\sqrt{9}}{2}\cot\theta

Hence the equation of the tangent at (8cos⁡θ,18sin⁡θ)\left( \sqrt{8}\cos\theta,\sqrt{18}\sin\theta \right)is y−18sin⁡θ=−92cot⁡θ(x−8cos⁡θ)y - \sqrt{18}\sin\theta = - \frac{\sqrt{9}}{2}\cot\theta\left( x - \sqrt{8}\cos\theta \right)

Therefore, the tangent cuts the coordinates axes at the points (0,18sin⁡θ)\left( 0,\frac{\sqrt{18}}{\sin\theta} \right)and(8cos⁡θ,0)\left( \frac{\sqrt{8}}{\cos\theta},0 \right)

Thus the area of the triangle formed by this tangent and the coordinate axes is

A = 1218.8\frac{1}{2}\sqrt{18}.\sqrt{8}.

1cos⁡θsin⁡θ=6cos⁡θsin⁡θ=12cos⁡ec2θ\frac{1}{\cos\theta\sin\theta} = \frac{6}{\cos\theta\sin\theta} = 12\cos ec2\theta

But cosec2θ is smallest when θ =π4\frac{\pi}{4}. Therefore A is smallest when θ = π4\frac{\pi}{4}.

Hence the required point is

(8,12,18.12)=(2,3)\left( \sqrt{8},\frac{1}{\sqrt{2}},\sqrt{18}.\frac{1}{\sqrt{2}} \right) = (2,3)