6.7 Q-4

Question Statement

Find the equations of the normal to the parabola $y^2 = 8x$ which are parallel to the line $2x + 3y = 10$ .

Background and Explanation

To solve this problem, we need to find the equation of the normal to the given parabola that is parallel to the given line. The key steps are:

Find the slope of the normal: The normal to a parabola at a given point is perpendicular to the tangent. The slope of the tangent at any point on a parabola can be found by differentiating the equation of the parabola.
Set the slope of the normal equal to the slope of the given line: Since the normal line is parallel to the given line, their slopes will be the same.
Use the point-slope form of the equation of a line to find the equation of the normal at the given point.

Solution

Step 1: Differentiate the equation of the parabola

The given equation of the parabola is:

y^2 = 8x \tag{1}

Differentiate both sides with respect to $x$ :

2y \frac{dy}{dx} = 8

Solving for $\frac{dy}{dx}$ :

\frac{dy}{dx} = \frac{8}{2y} = \frac{4}{y}

Thus, the slope of the tangent at any point on the parabola is $\frac{4}{y}$ .

Step 2: Find the slope of the normal

The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal is:

\text{Slope of normal} = -\frac{y}{4}

Step 3: Set the slope of the normal equal to the slope of the given line

The given line is:

2x + 3y = 10

To find the slope of this line, we rewrite it in slope-intercept form $y = mx + b$ . Solving for $y$ :

3y = -2x + 10 \quad \Rightarrow \quad y = -\frac{2}{3}x + \frac{10}{3}

Thus, the slope of the given line is $-\frac{2}{3}$ .

Since the normal to the parabola is parallel to the given line, their slopes must be equal. Therefore, we set the slope of the normal equal to $-\frac{2}{3}$ :

-\frac{y}{4} = -\frac{2}{3}

Solving for $y$ :

\frac{y}{4} = \frac{2}{3} \quad \Rightarrow \quad y = \frac{8}{3}

Step 4: Find the corresponding $x$ -coordinate

Substitute $y = \frac{8}{3}$ into the original equation of the parabola $y^2 = 8x$ :

\left(\frac{8}{3}\right)^2 = 8x

\frac{64}{9} = 8x \quad \Rightarrow \quad x = \frac{8}{9}

Thus, the point on the parabola where the normal is parallel to the given line is $\left( \frac{8}{9}, \frac{8}{3} \right)$ .

Step 5: Find the equation of the normal

Now, using the point $\left( \frac{8}{9}, \frac{8}{3} \right)$ and the slope $-\frac{2}{3}$ (since the normal is parallel to the given line), we use the point-slope form of the equation of a line:

y - y_1 = m(x - x_1)

Substitute $m = \frac{2}{3}$ , $x_1 = \frac{8}{9}$ , and $y_1 = \frac{8}{3}$ :

y - \frac{8}{3} = \frac{2}{3} \left( x - \frac{8}{9} \right)

Simplifying:

y - \frac{8}{3} = \frac{2}{3}x - \frac{16}{27}

y = \frac{2}{3}x - \frac{16}{27} + \frac{8}{3}

Convert $\frac{8}{3}$ to have a denominator of 27:

y = \frac{2}{3}x + \frac{72}{27} - \frac{16}{27} = \frac{2}{3}x + \frac{56}{27}

Multiply through by 27 to eliminate the denominator:

27y = 18x + 56

Rearranging into standard form:

18x - 27y + 56 = 0

Thus, the equation of the normal is:

18x - 27y + 56 = 0

Key Formulas or Methods Used

Differentiation: To find the slope of the tangent line to the parabola at a given point.

\frac{dy}{dx} = \text{Slope of tangent}

Negative Reciprocal: The slope of the normal is the negative reciprocal of the tangent’s slope:

\text{Slope of normal} = -\frac{1}{\text{Slope of tangent}}

Point-Slope Form: To write the equation of the normal:

y - y_1 = m(x - x_1)

where $m$ is the slope and $(x_1, y_1)$ is the point on the line.

Summary of Steps

Differentiate the equation of the parabola to find the slope of the tangent.
Find the slope of the normal by taking the negative reciprocal of the tangent’s slope.
Set the slope of the normal equal to the slope of the given line.
Solve for $y$ , then substitute into the original equation of the parabola to find the corresponding $x$ -coordinate.
Use the point-slope form of the equation of a line to find the equation of the normal.
Simplify the equation into standard form.