Question

What is the equation of the normal line to the parabola $y = {x^2} - 5x + 1$that is parallel to the line $x - 3y = 5$ ?

Answer 1

We know that if two lines are parallel, then the slopes of these two lines are equal. Therefore, we can find the slope of the required line using the given question statement. Also, we know that if we can find the slope of tangent to the curve at a point, then we can find the slope of normal at that point using the relation ${m_1}{m_2} = - 1$.

Complete step-by-step answer:

In the above question,
The equation of parabola is given as $y = {x^2} - 5x + 1$ and the equation of line is given as $x - 3y = 5$.
It is also given as that required line is normal to the parabola and parallel to the given line.
Therefore, the slope of the required line is equal to the given line.
Now, we will find the slope of the given line, that is $x - 3y = 5$.
We know that, slope of a line $= \dfrac{{ - coefficient\,of\,x}}{{coefficient\,of\,y}}$
Now, substitute the values from the equation of line in the above equation.
Slope of a line $= \dfrac{{ - 1}}{{ - 3}}$
$\Rightarrow \dfrac{1}{3}$
Now, we know that the slope of a tangent and a normal to a curve follows a relation as ${m_1}{m_2} = - 1$.
Now, we will find the slope of tangent to the parabola.
We have,
$y = {x^2} - 5x + 1$
Now, we will use a differentiation method to find the slope of the tangent.
Differentiate both sides,
$\Rightarrow \dfrac{{dy}}{{dx}} = 2x - 5$
Using, ${m_1}{m_2} = - 1$ we will find the slope of normal.
$\left( {2x - 5} \right){m_2} = - 1$
$\Rightarrow {m_2} = \dfrac{1}{{5 - 2x}}$
Now, we can evaluate $\dfrac{1}{{5 - 2x}}$ with $\dfrac{1}{3}$ to find the x-coordinate of the point of intersection.
$\dfrac{1}{{5 - 2x}} = \dfrac{1}{3}$
$\Rightarrow 2x = 5 - 3$
$\Rightarrow 2x = 2$
$\Rightarrow x = 1$
Now, put the above value of x in the equation of parabola to find the y-coordinate of the point of intersection.
$y = {1^2} - 5 \times 1 + 1$
$y = 2 - 5$
$y = - 3$
Now, we will use the point slope form to find the equation of the required line.
$y - {y_1} = m\left( {x - {x_1}} \right)$
Now, put the values in the above equation as ${y_1} = - 3$, $x = 1$ and $m = \dfrac{1}{3}$.
$\Rightarrow y - \left( { - 3} \right) = \dfrac{1}{3}\left( {x - 1} \right)$
$\Rightarrow y + 3 = \dfrac{1}{3}\left( {x - 1} \right)$
$\Rightarrow y = \dfrac{1}{3}\left( {x - 1} \right) - 3$
$\Rightarrow y = \dfrac{x}{3} - \dfrac{{10}}{3}$
Therefore, the equation of the required line is $y = \dfrac{x}{3} - \dfrac{{10}}{3}$.

Note: In the above question, we can also use the intercept form to find the equation of line. But for that we have to first find the intercept of the line on the y-axis using a given point and the value of slope. We can also find the point of intersection of the line and the curve by solving them simultaneously.