Question

The mirror image of the parabola ${y^2} = 4x$ in the tangent to the parabola at the point $\left( {1,2} \right)$ is
A. ${\left( {x - 1} \right)^2} = 4\left( {y + 1} \right)$
B. ${\left( {x + 1} \right)^2} = 4\left( {y + 1} \right)$
C. ${\left( {x + 1} \right)^2} = 4\left( {y - 1} \right)$
D. ${\left( {x - 1} \right)^2} = 4\left( {y - 1} \right)$

Answer 1

In the above question, we are given a parabola ${y^2} = 4x$ . There is an another parabola which is mirror image of the given parabola, located at the tangent to the given parabola at the point $\left( {1,2} \right)$ . We have to find the equation of the other parabola. First, we have to find the equation of the tangent to the given parabola at $\left( {1,2} \right)$ . After that we have to point the mirror image of all the arbitrary points of the given parabola in the tangent.

Complete step by step answer:
Given parabola is,
$\Rightarrow {y^2} = 4x$
Therefore, when $y = 2t$then $x = {t^2}$ .
So all the arbitrary points of the given parabola are of the form $\left( {{t^2},2t} \right)$ .
Now we have to find the equation of the tangent at point $\left( {1,2} \right)$ .
Since, slope of the tangent of a curve is given by $\dfrac{{dy}}{{dx}}$
Hence, slope of the tangent of ${y^2} = 4x$ is given by,
$\Rightarrow \dfrac{{d{y^2}}}{{dx}} = \dfrac{{d4x}}{{dx}}$

That gives,
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4$
Therefore, slope of the tangent is
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{2y}}$
Now, slope of the tangent at point $\left( {1,2} \right)$ is
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = \dfrac{4}{{2{y_1}}}$
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = \dfrac{4}{{2 \cdot 2}}$
Hence,
$\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {1,2} \right)}} = 1$

Now equation of tangent at the point $\left( {1,2} \right)$ is given by
$\Rightarrow \left( {y - 2} \right) = \left( {x - 1} \right)$
That gives,
$\Rightarrow x - y + 1 = 0$
Now the image, say $\left( {h,k} \right)$ of the point $\left( {{t^2},2t} \right)$ in the tangent $x - y + 1 = 0$ is given by
$\Rightarrow \dfrac{{h - {t^2}}}{1} = \dfrac{{k - 2t}}{{ - 1}} = - \dfrac{{2\left( {{t^2} - 2t + 1} \right)}}{{1 + 1}}$
That gives,
$\Rightarrow h - {t^2} = 2t - k = - \left( {{t^2} - 2t + 1} \right)$
That gives,
$\Rightarrow h - {t^2} = - {t^2} + 2t - 1$ and $\Rightarrow k - 2t = {t^2} - 2t + 1$
Or we can write it as,
$\Rightarrow h = 2t - 1$ and $\Rightarrow k = {t^2} + 1$

Now, since $h = 2t - 1$ then
$\Rightarrow t = \dfrac{{h + 1}}{2}$
Therefore, substituting $t = \dfrac{{h + 1}}{2}$ in $k = {t^2} + 1$ , we can write
$\Rightarrow k = {\left( {\dfrac{{h + 1}}{2}} \right)^2} + 1$
That gives us,
$\Rightarrow k - 1 = \dfrac{{{{\left( {h + 1} \right)}^2}}}{4}$
Hence, we get
$\Rightarrow 4\left( {k - 1} \right) = {\left( {h + 1} \right)^2}$
Now, taking locus of all the arbitrary points of the mirror image of parabola, we can write its equation as
$\Rightarrow 4\left( {y - 1} \right) = {\left( {x + 1} \right)^2}$
$\therefore {\left( {x + 1} \right)^2} = 4\left( {y - 1} \right)$
That is the required equation of the other parabola. Therefore, the mirror image of the parabola ${y^2} = 4x$ in the tangent to the parabola at the point $\left( {1,2} \right)$ is ${\left( {x + 1} \right)^2} = 4\left( {y - 1} \right)$ .

Hence, the correct option is C.

Note: The standard form of an equation of a parabola is ${y^2} = 4ax$ , where $a > 0$. The equation ${y^2} = 4ax$ represents the equation of a parabola whose coordinates of the vertex is at $\left( {0,{\text{ }}0} \right)$. The coordinates of the focus are $\left( { - {\text{ }}a,{\text{ }}0} \right)$.
The equation of directrix is $x{\text{ }} = {\text{ }}a$ or $x{\text{ }} - {\text{ }}a{\text{ }} = {\text{ }}0$ .
The equation of the axis is $y{\text{ }} = {\text{ }}0$ , the axis is along the positive x-axis.The length of its latus rectum is $4a$ and the distance between its vertex and focus is $a$.