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

Hint: 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 \[a\].