Let S be the circle in the XY-plane defined the equation x2 + y2 = 4. Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

Ellipse PYQ — JEE Advanced 2018 · Shift 1 · Q34 with Solution

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Correct answer: D

The circle is $x^2+y^2=4,$ so a point $P$ in the first quadrant can be written as $P=(2\cos\theta,\,2\sin\theta), \qquad 0<\theta<\frac{\pi}{2}.$

For the circle $x^2+y^2=4$ , the tangent at $(x_1,y_1)$ is $xx_1+yy_1=4.$

At $P=(2\cos\theta,2\sin\theta)$ , this becomes $2x\cos\theta+2y\sin\theta=4,$ or $x\cos\theta+y\sin\theta=2.$

On the $x$ -axis, $y=0$ : $x\cos\theta=2 \implies x=\frac{2}{\cos\theta}=2\sec\theta.$ So, $M=(2\sec\theta,0).$
On the $y$ -axis, $x=0$ : $y\sin\theta=2 \implies y=\frac{2}{\sin\theta}=2\csc\theta.$ So, $N=(0,2\csc\theta).$

Let the midpoint be $Q=(h,k)$ . Then

\qquad k=\frac{0+2\csc\theta}{2}=\csc\theta.$$ So, $$Q=(\sec\theta,\csc\theta).$$ 5. **Eliminate the parameter** We use the identity $$\cos^2\theta+\sin^2\theta=1.$$ Since $$\cos\theta=\frac{1}{h}, \qquad \sin\theta=\frac{1}{k},$$ substitute into the identity: $$\frac{1}{h^2}+\frac{1}{k^2}=1.$$ Multiplying by $h^2k^2$, $$h^2+k^2=h^2k^2.$$ Replacing $(h,k)$ by $(x,y)$, the locus is $$x^2+y^2=x^2y^2.$$ 6. **Match with options** This is exactly **Option D**. --- ### Verification with stored answer Stored correct answer: **D** Our derived answer: **D** So, the derived answer agrees with the stored answer.