JEE Main 2023 — Parabola Question with Solution

Given,

The equations of sides $AB$ and $AC$ of a triangle $ABC$ and have orthocentre at $\left(1,2\right)$

So, $AB\equiv \left(\lambda +1\right)x+\lambda y=4 ........\left(1\right)$

And $AC\equiv \lambda x+\left(1-\lambda \right)y+\lambda =0 ......\left(2\right)$ 

Also given vertex $A$ is on the $y-$axis

So, putting the value $x=0$ in equation $\left(1\right) \& \left(2\right)$ we get,

$y=\frac{4}{\lambda } \& y=\frac{\lambda }{\lambda -1}$

$\Rightarrow \frac{4}{\lambda }=\frac{\lambda }{\lambda -1}$

$\Rightarrow \lambda =2$

Now putting the value of $\lambda$ in both equation we get,

$AB\equiv 3x+2y=4 ........\left(3\right)$

$AC\equiv 2x-y+2=0 ......\left(4\right)$

Now solving above equation we get, $A\left(0,2\right)$

Now let $C\left(\alpha ,2\alpha +2\right)$ {using equation $\left(4\right)$}

Now we know that slopes of perpendicular lines is $-1$,

So, from diagram slope of altitude through $C\times \left(\text{slope of }AB\right)$ we get,

$\frac{2\alpha }{\alpha -1}\times \left(\frac{-3}{2}\right)=-1$

$\Rightarrow \alpha =\frac{-1}{2}$

Hence, point $C$ will be $C\left(\frac{-1}{2},1\right)$

Now let the equation of tangent be $y=mx+\frac{3}{2m}$ which passes through point $C\left(\frac{-1}{2},1\right)$

So, $1=m\left(\frac{-1}{2}\right)+\frac{3}{2m}$

$\Rightarrow m^2+2m-3=0$

$\Rightarrow m=1, -3$

Also we know that point of contact of tangent to parabola is given by $\left(\frac{a}{m^2},\frac{2a}{m}\right)\equiv \left(\frac{3}{2},3\right)$

Hence, $CT=\sqrt{{\left(\frac{3}{2}+\frac{1}{2}\right)}^2+{\left(3-1\right)}^2}=\sqrt{8}=2\sqrt{2}$