Question

A straight line parallel to the line \( 2x - y + 5 = 0 \) is also a tangent to the curve \( y^2 = 4x + 5 \). Then the point of contact is:

• A. \( (2, 1) \)
• B. \( (-1, 1) \)
• C. \( (1, 3) \)
• D. \( (3, 4) \)
• E. \( (-1, 2) \)

Hint:

The slope of a curve at any point $(x, y)$ is simply the derivative $dy/dx$. Parallel = Same Slope; Perpendicular = Negative Reciprocal Slope.

Answer 1

The Correct Option is B

Concept: Parallel lines have the same slope. We find the slope of the given line, set it equal to the derivative (\( \frac{dy}{dx} \)) of the curve at the point of contact \( (x_1, y_1) \), and solve for the coordinates.

Step 1: Find the required slope.
The line is \( 2x - y + 5 = 0 \), which can be written as \( y = 2x + 5 \). The slope \( m = 2 \). Since the tangent is parallel, the slope of the tangent at point \( (x_1, y_1) \) is also \( 2 \).

Step 2: Differentiate the curve.
Curve: \( y^2 = 4x + 5 \) Differentiating both sides with respect to \( x \): \[ 2y \frac{dy}{dx} = 4 \] \[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

Step 3: Solve for the point of contact.
Set the derivative equal to the slope: \[ \frac{2}{y_1} = 2 \quad \Rightarrow \quad y_1 = 1 \] Substitute \( y_1 = 1 \) back into the curve equation to find \( x_1 \): \[ (1)^2 = 4x_1 + 5 \] \[ 1 = 4x_1 + 5 \] \[ 4x_1 = -4 \quad \Rightarrow \quad x_1 = -1 \] The point of contact is \( (-1, 1) \).