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Question: From the point (15, 12), three normals are drawn to the parabola \({{\text{y}}^2} = 4{\text{x}}\), t...

From the point (15, 12), three normals are drawn to the parabola y2=4x{{\text{y}}^2} = 4{\text{x}}, then centroid of triangle formed by three co-normal points is
A. (163,0) B. (4,0) C. (263,0) D. (6,0)  {\text{A}}{\text{. }}\left( {\dfrac{{16}}{3},0} \right) \\\ {\text{B}}{\text{. }}\left( {4,0} \right) \\\ {\text{C}}{\text{. }}\left( {\dfrac{{26}}{3},0} \right) \\\ {\text{D}}{\text{. }}\left( {6,0} \right) \\\

Explanation

Solution

Hint : To find the point of centroid of the triangle formed by the three co-normal points, we first write the equation of a normal to the given parabola using the general equation of a normal to a parabola. Using the general form of a point of the normal we deduce the three co-normal points and compute the centroid using its formula.

Complete step by step solution :
Given Data,
Point (15, 12)
Equation of parabola y2=4x{{\text{y}}^2} = 4{\text{x}}

We know the general equation of a normal from a point to a parabola given by the equation, y2=4ax{{\text{y}}^2} = 4{\text{ax}} is given by, y = - xt + 2at + at3{\text{y = - xt + 2at + a}}{{\text{t}}^3} and the general form of the point from which the normal comes is of the form P(t2,2t){\text{P}}\left( {{{\text{t}}^2},2{\text{t}}} \right).

Given the equation of the parabola is y2=4x{{\text{y}}^2} = 4{\text{x}}
By comparing it with the general equation of the parabola we get, a = 1.

Therefore the equation of the normal to the parabola, y2=4x{{\text{y}}^2} = 4{\text{x}} is given by
y = - xt + 2at + at3⇒y = - xt + 2t + t3{\text{y = - xt + 2at + a}}{{\text{t}}^3} \Rightarrow {\text{y = - xt + 2t + }}{{\text{t}}^3}
This equation passes through the point (15, 12), hence this point should satisfy this equation.
⇒12 = - 15t + 2t + t3\Rightarrow 12{\text{ = - 15t + 2t + }}{{\text{t}}^3}
⇒t3−13t - 12 = 0\Rightarrow {{\text{t}}^3} - 13{\text{t - 12 = 0}}
⇒(t + 1)(t2−t - 12) = 0\Rightarrow \left( {{\text{t + 1}}} \right)\left( {{{\text{t}}^2} - {\text{t - 12}}} \right){\text{ = 0}}
⇒t = - 1, - 3, 4\Rightarrow {\text{t = - 1, - 3, 4}}

Therefore the co-normal points can be determined using the general form of point, P(t2,2t){\text{P}}\left( {{{\text{t}}^2},2{\text{t}}} \right)
We get the co-normal points as (1, -2), (9, -6), (16, 8)

Now we know the formula of a centroid given three points (a, d), (b, e) and (c, f) is (a + b + c3,d + e + f3)\left( {\dfrac{{{\text{a + b + c}}}}{3},\dfrac{{{\text{d + e + f}}}}{3}} \right)
Therefore the centroid of the three co-normal points (1, -2), (9, -6), (16, 8) is (1 + 9 + 163, - 2 - 6 + 83)\left( {\dfrac{{{\text{1 + 9 + 16}}}}{3},\dfrac{{{\text{ - 2 - 6 + 8}}}}{3}} \right)
⇒(263,0)\Rightarrow \left( {\dfrac{{26}}{3},0} \right)

Therefore the centroid of triangle formed by three co-normal points is(263,0)\left( {\dfrac{{26}}{3},0} \right)
Option C is the correct answer.

Note : In order to solve this type of problems the key is to know the concept of a parabola and a normal draw to it from a point and the respective general equations of the parabola, the normal drawn to it and the point it comes from. We use all of these general forms to compute the three co-normal points formed.
Also we use the formula of a centroid of a triangle, so a good knowledge of triangle properties and their respective formulae is to be known.