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Question: Normals are drawn to the parabola from a point P with slopes \({m_1},{m_2},{m_3}\) to \({y^{^2}} = 4...

Normals are drawn to the parabola from a point P with slopes m1,m2,m3{m_1},{m_2},{m_3} to y2=4x{y^{^2}} = 4x. If the locus of m1m2=α{m_1}{m_2} = \alpha is a part of parabola itself then the value of α\alpha is
A) 1
B) 2
C) 2 or -2
D) -2

Explanation

Solution

First of all we have to understand the normal of the parabola which is the line perpendicular to the tangent of the parabola at the point of the contact we can also understand the normal with the help of the diagram given below:


As given in the diagram above normal is the line which is perpendicular to the tangent (is a line to a plane curve at a given point P is a straight line which just touches the curve at the point) and meets the tangent at point P.
A parabola is a curve where any point is at an equal distance from a fixed straight line (directrix) and a fixed point (focus) which can be understood by the diagram given below:

Now, the locus of the parabola is defined as the locus of a point that moves so that it always the same distance from the focus and a given line directrix.
So, first of all, we will compare the given parabola equation which is y2=4x{y^{^2}} = 4x with the equation of the parabola y2=4ax{y^2} = 4ax from here we can obtain the value of a so, to put it into the equation of normal which is y=mx−2am−am3y = mx - 2am - a{m^3}
After obtaining the equation of normal we will let that the normal passes through the points (h, k) and we will replace the points (x, y) with (h, k) and on solving the obtained equation and comparing it with the given equation of parabola which is y2=4x{y^{^2}} = 4x we can find the value of α\alpha .

Complete step by step answer:
Given,
Equation of parabola = y2=4x{y^{^2}} = 4x
Slopes of parabola are m1,m2,m3{m_1},{m_2},{m_3}
m1m2=α{m_1}{m_2} = \alpha
Step 1: First of all we will compare the equation of parabola as given in the question which is y2=4x{y^{^2}} = 4x with the equation of parabola y2=4ax{y^2} = 4ax to obtain the value of a.
Hence, on comparing the value of a = 1
Step 2: Now, as we know the equation of normal for the parabola y2=4ax{y^2} = 4ax is
y=mx−2am−am3y = mx - 2am - a{m^3}On substituting the value of a as obtained in the step 1 in the equation of normal.
y=mx−2m−m3y = mx - 2m - {m^3}
Step 3: Let, the normal passes through the points (h, k) we can also understand with the help of the diagram given below

On substituting the points (h, k) in the equation of normal,
⇒k=mh−2m−m3\Rightarrow k = mh - 2m - {m^3}
On rearranging the obtained equation of the normal,
m3+m(2−h)+k=0.......................(1){m^3} + m(2 - h) + k = 0.......................(1)
Step 4: Now, to find the slopes we will compare the equation (1) with the equation of the normal.
Hence, on comparing
⇒m1+m2+m3=da\Rightarrow {m_1} + {m_2} + {m_3} = \dfrac{d}{a}
Where, d = 0
⇒m1+m2+m3=0\Rightarrow {m_1} + {m_2} + {m_3} = 0……………………………(2)
And, m1m2+m2m3+m3m1=ba{m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1} = \dfrac{b}{a}
Where, b = (2-h) and a = 1
⇒m1m2+m2m3+m3m1=(2−h)\Rightarrow {m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1} = (2 - h)……………………….(3) and,
⇒m1m2m3=−ca\Rightarrow {m_1}{m_2}{m_3} = - \dfrac{c}{a}
Where, c = k and a = 1 hence,
m1m2m3=−k{m_1}{m_2}{m_3} = - k…………………………………………(4)
Step 5: on substituting the value of m1m2=α{m_1}{m_2} = \alpha in the equation (4) as given in the question,
⇒αm3=−k ⇒m3=−ka  \Rightarrow \alpha {m_3} = - k \\\ \Rightarrow {m_3} = \dfrac{{ - k}}{a} \\\
Step 6: On substituting the value of m3{m_3}in equation (1)
⇒(−ka)3−ka(2−h)+k=0 ⇒−k3a3−ka(2−h)+k=0  \Rightarrow {\left( {\dfrac{{ - k}}{a}} \right)^3} - \dfrac{k}{a}(2 - h) + k = 0 \\\ \Rightarrow \dfrac{{ - {k^3}}}{{{a^3}}} - \dfrac{k}{a}(2 - h) + k = 0 \\\
On solving the obtained equation,
⇒k2=α2h−2α2+α2\Rightarrow {k^2} = {\alpha ^2}h - 2{\alpha ^2} + {\alpha ^2}………………………………(5)
Step 7: On replacing the points (h, k) with the points (x, y) in the obtained equation (5)
⇒y2=α2x−2α2+α3\Rightarrow {y^2} = {\alpha ^2}x - 2{\alpha ^2} + {\alpha ^3}………………………………(6)
Step 8: On comparing the obtained equation (6) with
⇒y2=4x ⇒α2=4  \Rightarrow {y^2} = 4x \\\ \Rightarrow {\alpha ^2} = 4 \\\
And,
⇒−2α2+α3=0 ⇒α2(−2+α)=0 ⇒α=2  \Rightarrow - 2{\alpha ^2} + {\alpha ^3} = 0 \\\ \Rightarrow {\alpha ^2}( - 2 + \alpha ) = 0 \\\ \Rightarrow \alpha = 2 \\\

Hence, we have obtained the value α=2\alpha = 2 by solving the equation of slope and comparing it with the equation of parabola y2=4x{y^2} = 4x as given in the question. Therefore, the correct option is (B).

Note:
A tangent is a line to a plane curve at a given point is a straight line which just touches the curve at the point.
The derivative of a function at a given point is the slope of the tangent line at that point so, you Have to set the derivative of the parabola equal to the slope of the tangent.
To find the slope it is necessary to compare the equation of the normal with the equation of the parabola.