Question

The mirror image of the parabola $y^2=4x$ in the tangent to the parabola at the point $(1,2)$ is

• A. $(x-1{)}^2=4(y+1)$
• B. $(x+1{)}^2=4(y+1)$
• C. $(x+1{)}^2=4(y-1)$
• D. $(x-1{)}^2=4(y-1)$

Answer 1

The correct option is C $(x+1{)}^2=4(y-1)$

Given the equation of parabola is
$y^2=4x$

Here $a=1$

Let any point on the given parabola is $(t^2,2t)$.

Slope of tangent at $(1,2)$ $=(\frac{dy}{dx}{)}_{(1,2)}=1$

The equation of the tangent at $(1,2)$ is

$y=x+1$

or, $x-y+1=0$

The image $(h,k)$ of the point $(t^2,2t)$ in $x-y+1=0$ is given by

$\frac{h-t^2}{1}=\frac{k-2t}{-1}=-\frac{2(t^2-2t+1)}{1+1}$

$\Rightarrow h-t^2=-t^2+2t-1$ and $k-2t=t^2-2t+1$

$\Rightarrow h=2t-1$ and $k=t^2+1$

Substitute the value of $t$ from $h=2t-1$ in $k,$ we get
$(h+1{)}^2=4(k-1)$

The required equation of reflection is $(x+1{)}^2=4(y-1)$