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 = 44/9

Step-by-step solution

Identify the conditions for normal line

A line is normal to a parabola if it is perpendicular to the tangent at the point of contact. First, determine the slope of the tangent to the parabola at any point.

Find the derivative of the parabola equation

Differentiate the given parabola equation with respect to x. The given equation is: y^2 = 4x Differentiating both sides with respect to x gives: 2y(dy/dx) = 4 Solving for dy/dx gives: dy/dx = 2/y

Express the slope of the line

The line given is 2x - 3y - k = 0. Rewrite it in the slope-intercept form: y = (2/3)x - k/3 The slope of the line is 2/3.

Determine the slope of the normal line and set up the equation

Since the slope of the tangent at any point (x1, y1) on the parabola y^2 = 4x is 2/y1, the slope of the normal line, which is perpendicular to this tangent line, is the negative reciprocal of 2/y1, which gives: slope_normal = -y1/2. Since the normal line has slope 2/3, we set up the equation: -y1/2 = 2/3.

Solve for y1

Solve the equation -y1/2 = 2/3: y1 = -4/3.

Find the corresponding x1

Plug y1 = -4/3 back into the equation y^2 = 4x to find x1: (-4/3)^2 = 4x 16/9 = 4x x = 4/9.

Determine k

The point (4/9, -4/3) lies on the line 2x - 3y - k = 0. Substitute x and y into the line equation: 2(4/9) - 3(-4/3) - k = 0 8/9 + 4 - k = 0 8/9 + 36/9 - k = 0 44/9 - k = 0 k = 44/9.

Key concepts

slope of tangent

The slope of a tangent line to a curve at a particular point is a measure of how steep the line is at that precise location. To find the slope of the tangent to a parabola, we first differentiate the equation of the parabola. For example, given the equation of the parabola: \[ y^2 = 4x \],we differentiate both sides with respect to x to obtain the derivative. Differentiation is a process in calculus used to find the rate at which one quantity changes with respect to another. In this case, differentiating \[ y^2 = 4x \],with respect to x gives: \[ 2y \frac{dy}{dx} = 4 \].Solving for \( \frac{dy}{dx} \), which represents the slope of the tangent line to the parabola at any point \( (x, y) \), we get:\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \].Thus, the slope of the tangent to the parabola at any point \( (x, y) \) is \( \frac{2}{y} \).

differentiation

Differentiation is a fundamental concept in calculus that helps us determine the rate at which a function is changing at any given point. It's particularly useful in finding the slope of the tangent lines to curves. When differentiating an equation like \[ y^2 = 4x \],we look to find \( \frac{dy}{dx} \), which represents the slope of the curve at any point on the parabola. By differentiating both sides with respect to x, we apply the chain rule to get:\[ 2y \frac{dy}{dx} = 4 \],and then solve for \( \frac{dy}{dx} \) to obtain: \[ \frac{dy}{dx} = \frac{2}{y} \].This result shows how the y-coordinate of a point on the parabola affects the slope of the tangent line at that point. Remember, correct differentiation is key in understanding how functions behave and change.

geometry

Geometry plays a crucial role when analyzing curves and lines, especially in the context of this problem involving the normal to a parabola. A normal to a curve at a given point is a line perpendicular to the tangent at that point. To determine if a line is normal to a given parabola, we need to explore both slopes - that of the tangent and the given line. For example, the line provided in the problem is: \[ 2x - 3y - k = 0 \],which can be rewritten in slope-intercept form as: \[ y = \frac{2}{3}x - \frac{k}{3} \].Here, the slope of the line is \( \frac{2}{3} \). When considering a line normal to the parabola, the slope of the normal line is the negative reciprocal of the tangent's slope. Therefore, if the slope of the tangent at any point \( (x_1, y_1) \) on the parabola is \( \frac{2}{y_1} \), the slope of the normal is \[ -\frac{y_1}{2} \]. By solving the equation \[-\frac{y_1}{2} = \frac{2}{3} \], we determine \( y_1 \) and subsequently \( x_1 \) to find the correct value of k. Understanding these geometric relationships and the role of slopes helps in solving complex problems involving curves and their normals.