Question

If a point $P$ on the axis of the parabola $y^2=4x$ is taken such that the point is at shortest distance from the circle $x^2+y^2+2x-2\sqrt{2}y+2=0$.Common tangents are drawn to the circle and the parabola from $P$.If the area of the $\Delta PAB$ is $\sqrt{a}$ sq. units where $A$ and $B$ are the points of contact on two distinct tangents from $P$ on circle and parabola respectively, then $a=$

Answer 1

Axis of the parabola $y^2=4x$ is $y=0$, and the point which is at shortest distance from the circle.
$x^2+y^2+2x-2\sqrt{2}y+2=0$ is $(-1,0)$

Since, $(-1,0)$ lies on the directrix of the parabola hence tangents to the parabola are $y=x+1$ and $y=-x-1$ and these lines are also the tangents to the given circle. Here, $△PAB$ is possible two ways one shown by using $PAB$ and other by using $P{A}_1{B}_1$.
Length of $PB=P{B}_1=1$, also $A(1,2)$ and $A_1(1,-2)$ as $A$ and $A_1$ are the endpoints of the latusrectum of the parabola because latus rectum is only focal chord that subtends rightangle at axis of the parabola where it meets with directrix.
Here, $△PAB$ and $△P{A}_1{B}_1$ are right angle triangles.
So, area of $△PAB=\frac{1}{2}\times PA\times PB=\frac{1}{2}\times 1\times 2\sqrt{2}=\sqrt{2}\text{sq.unit}$
Also, area of
$\left(△P{A}_1{B}_1\right)=\frac{1}{2}\times P{A}_1\times P{B}_1=\frac{1}{2}\times 1\times 2\sqrt{2}=\sqrt{2}$ sq unit
Hence, area of $△PAB=\sqrt{2}$ sq unit