Question

If line \(2 x-3 y-k=0\) is normal to parabola \(y^{2}=4 x\), then \(k\) is equal to (1) \(\frac{22}{9}\) (2) \(\frac{2}{3}\) (3) \(\frac{44}{9}\) (4) \(\frac{4}{27}\)

Short answer

k = \(\frac{44}{9}\)

Step-by-step solution

- Understand the problem

Identify that the line given by the equation \(2x - 3y - k = 0\) is normal to the parabola \(y^2 = 4x\). This means the slope of the line is the negative reciprocal of the derivative of the parabola at the point of intersection.

- Find the slope of the line

Rewrite the line equation in slope-intercept form: \(y = \frac{2}{3}x - \frac{k}{3}\). Therefore, the slope of the line is \(\frac{2}{3}\).

- Find the derivative of the parabola

Differentiate \(y^2 = 4x\) with respect to \(x\). This gives: \(2y \frac{dy}{dx} = 4\), or \(\frac{dy}{dx} = \frac{2}{y}\).

- Set the normal slope condition

The line is normal to the parabola, so \(\frac{dy}{dx}\) at the point of intersection is the negative reciprocal of the line's slope: \(\frac{dy}{dx} = -\frac{3}{2}\).

- Solve for y

Set \(\frac{2}{y} = -\frac{3}{2}\) and solve for \(y\). This gives \(y = -\frac{4}{3}\).

- Find corresponding x

Substitute \(y = -\frac{4}{3}\) into \(y^2 = 4x\). This gives \(\left( -\frac{4}{3} \right)^2 = 4x\), so \(\frac{16}{9} = 4x\). Therefore, \(x = \frac{4}{9}\).

- Substitute values into the line equation

Substitute \(x = \frac{4}{9}\) and \(y = -\frac{4}{3}\) into the line equation \(2x - 3y - k = 0\) to find \(k\). This gives \(2 \left( \frac{4}{9} \right) - 3 \left( -\frac{4}{3} \right) - k = 0\).

- Solve for k

Simplify the equation: \(\frac{8}{9} + 4 - k = 0\). Combine like terms to find: \(\frac{8}{9} + \frac{36}{9} - k = 0\), so \(\frac{44}{9} - k = 0\). Thus, \(k = \frac{44}{9}\).

Key concepts

Slope of a Line

The slope of a line is a measure of its steepness and direction. In this context, the slope is given by putting the line equation into slope-intercept form. Given the line equation: \(2x - 3y - k = 0\), we can rewrite it as: \(y = \frac{2}{3}x - \frac{k}{3}\). Here, the slope is the coefficient of \(x\), which is \(\frac{2}{3}\). Understanding the slope is essential since it helps us determine the relationship between the line and the parabola.
It's important to note that a line being normal to a curve means it is perpendicular to the tangent of the curve at the point of intersection. The slope of the normal line is the negative reciprocal of the tangent slope at that point.

Derivative Calculation

The derivative of a function gives us the slope of the tangent to the curve at any point. For the parabola \(y^2 = 4x\), we differentiate with respect to \(x\) to find the slope of the tangent. Differentiating implicitly, we have:
\[2y \frac{dy}{dx} = 4\]
Simplifying, we get:
\[\frac{dy}{dx} = \frac{2}{y}\]
This derivative tells us the slope of the tangent to the parabola at any point \((x, y)\). To find where the line is normal to the parabola, we need the derivative at the point of intersection to match the negative reciprocal of the line's slope, which is: \[-\frac{3}{2}\]. By equating \[\frac{dy}{dx}\] to \[-\frac{3}{2}\], we can solve for \(y\):
\[\frac{2}{y} = -\frac{3}{2}\]
Solving for \(y\), we find:
\[y = -\frac{4}{3}\]

Normal to a Parabola

A line normal to a parabola crosses the parabola at a right angle to the tangent at the point of intersection. After determining the proper \(y\)-value from the derivative, we next find the corresponding \(x\)-value:
Given \(y = -\frac{4}{3}\), substituting into the equation of the parabola \(y^2 = 4x\), we get:
\[\left( -\frac{4}{3} \right)^2 = 4x\]
Simplifying, we find:
\[\frac{16}{9} = 4x\]
So, \[x = \frac{4}{9}\]
We now substitute these values \(x = \frac{4}{9}\) and \(y = -\frac{4}{3}\) into the line equation \(2x - 3y - k = 0\) to solve for \(k\):
Substituting, we get:
\[2 \left( \frac{4}{9} \right) - 3 \left( -\frac{4}{3} \right) - k = 0\]
This simplifies to:
\[\frac{8}{9} + 4 - k = 0\]
Combining like terms, we have:
\[\frac{44}{9} - k = 0\]
Solving for \(k\), we find:
\[k = \frac{44}{9}\]
Thus, the value of \(k\) ensuring the line is normal to the parabola is \(\frac{44}{9}\).