JEE Main 2019 — Parabola Question with Solution

The equation of the tangent to the parabola $y^2=4ax$ at $\left(x_1, y_1\right)$ is $y{y}_1=2a\left(x+x_1\right).$ 

Equation of tangent to the parabola $y^2=4x$ at $\left(1, 2\right)$ is
$y\cdot 2=2\left(x+1\right)$

$\Rightarrow x-y+1=0 ...\left(i\right)$

The equation of the family of circles touching a line $l=0$ at $\left(x_1, y_1\right)$ is ${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+\lambda l=0.$

Equation of family of circle touching the parabola at $\left(1, 2\right)$ is

${\left(x-1\right)}^2+{\left(y-2\right)}^2+\lambda \left(x-y+1\right)=0 ...\left(ii\right)$

Since circle touches $x$-axis, hence by putting $y=0,$ we get

$x^2+\left(\lambda -2\right)x+\left(\lambda +5\right)=0$

As the circle touches the $x$-axis, hence this equation will have only one root.

And we know that a quadratic equation$a{x}^2+bx+x=0, a\ne 0,$  has only one root when its discriminant $D=b^2-4ac$ is zero.

Now $D=0$

$\Rightarrow {\left(\lambda -2\right)}^2-4\cdot 1\cdot \left(\lambda +5\right)=0$

$\Rightarrow {\lambda }^2-4\lambda +4-4\lambda -20=0$

$\Rightarrow {\lambda }^2-8\lambda -16=0$

Now, applying the Sridharacharya formula, we get

$\lambda =\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\times 1\times \left(-16\right)}}{2\times 1}$

$\Rightarrow \lambda =4\pm 4\sqrt{2}$ for smaller circle $\lambda =4-4\sqrt{2}.$

Hence, the equation of the circle is ${\left(x-1\right)}^2+{\left(y-2\right)}^2+\left(4-4\sqrt{2}\right)\left(x-y+1\right)=0$

$\Rightarrow x^2+y^2-2x\left(1-2\sqrt{2}\right)-2y\left(4-2\sqrt{2}\right)+\left(9-4\sqrt{2}\right)=0$

Since, the circle touches the $x$-axis, hence the absolute value of the $y$-co-ordinate of the circle is the radius of the circle.

Thus, the radius of smaller circle $r=4-2\sqrt{2}$ units.

Area of circle $\pi r^2=\pi {\left(4-2\sqrt{2}\right)}^2$

$=\pi \left(16-16\sqrt{2}+8\right)=\pi \left(24-16\sqrt{2}\right)$

$=8\pi \left(3-2\sqrt{2}\right)$ sq units.