VerifiedHint: To solve this question we will first of all determine the equation of normal of the given parabola. The equation of normal of parabola of type, \[{{y}^{2}}=4x\] an point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given by,
\[\left( y-{{y}_{1}} \right)=\dfrac{-1}{\dfrac{dy}{dx}}\left( x-{{x}_{1}} \right)\]
After obtaining the equation of normal and assuming co – ordinates of point P we will try to determine the value of slope m of normal. Thus, it will help us to get the value of co – ordinates of point P.
Complete step-by-step solution:
Let us assume the point on the parabola is P.
We will first assume the coordinates of P.
Let x – coordinate of P be \[{{m}^{2}}\], then as P lies on the parabola \[{{y}^{2}}=4x\]. So, it must satisfy the equation \[{{y}^{2}}=4x\].
Substituting \[x={{m}^{2}}\] as \[{{y}^{2}}=4x\] to get y – coordinate of P we get,
\[{{y}^{2}}=4\left( {{m}^{2}} \right)\]
Taking square roots on both sides we get,
\[y=\pm 2m\]
So we can consider the y – coordinate of P as +2m or -2m.
Let it be -2m.
When P = \[\left( {{m}^{2}},-2m \right)\]
So, we have a figure as,
Now we have to consider normal at P.
The equation of normal of parabola of type, \[{{y}^{2}}=4x\] an point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given by,
\[\left( y-{{y}_{1}} \right)=\dfrac{-1}{\dfrac{dy}{dx}}\left( x-{{x}_{1}} \right)\]
Given \[{{y}^{2}}=4x\], we will calculate \[\dfrac{dy}{dx}\] now,
Differentiating above equation with respect to x we get,
\[2y\dfrac{dy}{dx}=4\]
Dividing by 2y both sides,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{2}{y}\]
Here point P has \[\left( {{m}^{2}},-2m \right)\] as co – ordinate.
Then at P; \[\dfrac{dy}{dx}=\dfrac{2}{\left( -2m \right)}\] as y = 2m at P.
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{m}\]
Substituting \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{m}^{2}},-2m \right)\] and \[\dfrac{dy}{dx}=-\dfrac{1}{m}\] in equation of normal of parabola we get;
\[y-\left( -2m \right)=\dfrac{-1}{\dfrac{-1}{m}}\left( x-{{m}^{2}} \right)\]
Cancelling common negative,
$ \Rightarrow y+2m=m\left( x-{{m}^{2}} \right) $
$ \Rightarrow y+2m=mx-{{m}^{3}} $
Subtracting 2m both sides,
\[\Rightarrow y=mx-{{m}^{3}}-2m\] - (1)
This is the equation of normal.
Given that normal makes an equal angle with the axis.
The slope = m = \[\tan \dfrac{\pi }{4}\].
And the value of \[\tan \dfrac{\pi }{4}\] = 1.
\[\Rightarrow m=1\]
Substituting m = 1 in equation (1) we get,
$ \Rightarrow y=x-1-2 $
$ \Rightarrow y=x-3 $
Now finally we have to calculate \[P=\left( {{m}^{2}},-2m \right)\],
\[\Rightarrow P=\left( +1,-2 \right)\]
So the point is (1, -2) and it is (1, +2) when P is taken as (-y, +2m).
So option (a) is correct.
Note: Student may get confused while assuming co – ordinates of point P at \[\left( {{m}^{2}},+2m \right)\] or \[\left( {{m}^{2}},-2m \right)\]. Both are correct, you can proceed for selecting any one of above as co – ordinate of P and then proceed for solution. Finally at the end you can use the other left one to get the full solution.
