Question

Normals at two points $\left( {{x_1},{y_2}} \right)$ and $\left( {{x_2},{y_2}} \right)$ of parabola ${y^2} = 4x$ meet again on the parabola where ${x_1} + {x_2} = 4$ , then $\left| {{y_1} + {y_2}} \right|$ equals to
A. $\sqrt 2 $
B. $2\sqrt 2 $
C. $4\sqrt 2 $
D. None of these

Answer 1

Hint: Normals at a point on a parabola are lines perpendicular to the tangent of the parabola at the point of contact. The slope of the normal is the negative inverse of the derivative at that point as the derivative at that point is the slope of the tangent and normal is perpendicular to the tangent. We will first find the derivative of the function and find the slope of the normal then. Then we will use the formula for the normal of the parabola with point form.

Formula used: The formula used in this situation is the formula for the equation of normal at a point to a parabola. The equation of normal to the parabola ${y^2} = 4ax$ at a point $({x_1},{y_1})$ is given by:
$\left( {y - {y_1}} \right) = - \dfrac{{{y_1}}}{{2{x_1}}}\left( {x - {x_1}} \right)$
Where $ - \dfrac{{{y_1}}}{{2{x_1}}}$ is the slope of the normal and is the negative inverse of the derivative of the function. To find the equation of such a normal, we need a singular point and the function to find the slope by differentiating the function.
Alternatively, when the normal at a point $a{t_1}$ meets the parabola again at the point $a{t_2}$ then $a{t_2} = - a{t_1} - \dfrac{2}{{a{t_1}}}$

Complete step-by-step answer:
We can assume point P to be $P({x_1},{y_1}) \equiv (a{t_1}^2,2a{t_1})$ .This the parametric form of the points on a parabola. This point meets the parabola at a point $R(a{t^2},2at)$ . This implies that
$at = - a{t_1} - \dfrac{2}{{a{t_1}}}$
$t = - {t_1} - \dfrac{2}{{{t_1}}}$ $...\left( 1 \right)$ As
It is also given that point $Q\left( {{x_2},{y_2}} \right) \equiv (a{t_2}^2,2a{t_2})$ meets the parabola at $R(a{t^2},2at)$ . So,
$at = - a{t_2} - \dfrac{2}{{a{t_2}}}$
$t = - {t_2} - \dfrac{2}{{{t_2}}}$ $...\left( 2 \right)$
From the above two equations,
\[ - {t_1} - \dfrac{2}{{{t_1}}} = - {t_2} - \dfrac{2}{{{t_2}}}\]
$ \Rightarrow {t_1}{t_2} = 2$
It is given in the question that ${x_1} + {x_2} = 4$ therefore,
$a{t_1}^2 + a{t_2}^2 = 4$
$ \Rightarrow {t_1}^2 + {t_2}^2 = 4$ since $a = 1$ from the equation of the parabola
\[ \Rightarrow {t_1}^2 + {t_2}^2 + 2{t_1}{t_2} - 2{t_1}{t_2} = 4\]
\[ \Rightarrow {({t_1} + {t_2})^2} = 4 + 2{t_1}{t_2}\]
\[ \Rightarrow {({t_1} + {t_2})^2} = 4 + 4\]
\[ \Rightarrow \left| {{t_1} + {t_2}} \right| = \sqrt 8 = 2\sqrt 2 \]
The question asks for
$\left| {{y_1} + {y_2}} \right| = \left| {2{t_1} + 2{t_2}} \right|$
$ \Rightarrow 2\left| {{t_1} + {t_2}} \right| = 2\left( {2\sqrt 2 } \right) = 4\sqrt 2 $

Therefore, the correct option is $\left( c \right)4\sqrt 2 $

Note:
This question can also be solved in the classic method using the point form of the parabolic chords but there’s scope for mistakes in that method. This method is easier and requires less work. It gives the correct answer in fewer steps.