Question

The point of the curve ${y^2} = 2(x - 3)$ , at which the normal is parallel to the line $y - 2x + 1 = 0$ is
A. $(5,2)$
B. $\left( - \dfrac{1}{2}, - 2\right)$
C. $(5, - 2)$
D. $\left(\dfrac{3}{2},2\right)$

Answer 1

Hint: First, the slope of the normal at the required point is calculated in terms of assumed variables and then equated with the given numerical data to find the coordinates of the point.

Complete step by step solution:
We have been given the equation of the curve ${y^2} = 2(x - 3)$
Let, $({x_1},{y_1})$ be a point on the curve, at which, the normal to the curve is parallel to the given line $y - 2x + 1 = 0$
First, we will find the slope of the tangent to the curve at the point $({x_1},{y_1})$ .
We will find out the first derivative of the given equation of the curve
${y^2} = 2(x - 3)\\ \Rightarrow 2y \times \dfrac{{dy}}{{dx}} = 2\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{2y}}\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{y}$
So, the slope of the tangent to the curve in the form of a variable is $\dfrac{1}{y}$.
At the point $({x_1},{y_1})$, the slope will be $\dfrac{1}{{{y_1}}}$ .
Let, ${m_1} = \dfrac{1}{{{y_1}}}$
And ${m_2} = $ the slope of the normal to the curve at the point $({x_1},{y_1})$
As the product of the slopes of two perpendicular line is equal to $ - 1$
Then, ${m_1} \times {m_2} = - 1\\ \Rightarrow \dfrac{1}{{{y_1}}} \times {m_2} = - 1\\ \Rightarrow {m_2} = - {y_1}$
So, the slope of the normal to the curve at the point $({x_1},{y_1})$ is $ - {y_1}$.
It is given that the above normal is parallel to the line $y - 2x + 1 = 0$.
We will find the slope of the above line after writing it in the slope-intercept form $y = mx + c$.
$y - 2x + 1 = 0\\ \Rightarrow y = 2x - 1$
Comparing it with the slope-intercept form, we have
$m = 2$ i.e. the slope of the line $y - 2x + 1 = 0$ is $2$.
Since, the slopes of two parallel lines are equal, the slope of the normal will also be $2$.
So, $ - {y_1} = 2\\ \Rightarrow {y_1} = - 2$ (As slope of the normal in terms of assumed variable is $ - {y_1}$)
Putting the value of ${y_1}$ in the equation of the curve, we have
${y^2} = 2(x - 3)\\ \Rightarrow {({y_1})^2} = 2({x_1} - 3)\\ \Rightarrow 4 = 2{x_1} - 6\\ \Rightarrow 2{x_1} = 10\\ \Rightarrow {x_1} = 5$
So, $({x_1},{y_1})$ i.e. $(5, - 2)$ is the required point on the curve.

Option ‘C’ is correct

Note: When the slopes of two lines are equal, they will be parallel and never intersect each other, but, to be perpendicular, the product of the slopes of the two lines should be $ - 1$ . The equation of the tangent as well as the normal to a curve at a point on the curve can be found out by applying this concept.