Question

Equation of the line touching both parabolas \( y^2 = 4x \) and \( x^2 = -32y \) is:

• A. \( x + 2y + 4 = 0 \)
• B. \( 2x + y - 4 = 0 \)
• C. \( x - 2y - 4 = 0 \)
• D. \( x - 2y + 4 = 0 \)

Hint:

For a common tangent to two parabolas, derive the tangents using their respective standard forms and equate slopes to find a consistent line equation.

Answer 1

The Correct Option is D

Step 1: Given Parabolas and Their Standard Forms The given parabolas are:
1. \( y^2 = 4x \), which represents a rightward-opening parabola.
2. \( x^2 = -32y \), which represents a downward-opening parabola.
Step 2: Slopes of the Common Tangent
The general equation of a parabola \( y^2 = 4ax \) has a standard tangent equation: \[ yy_1 = 2a(x + x_1). \] For \( y^2 = 4x \), we have \( a = 1 \), so the equation of a tangent at \( (x_1, y_1) \) is: \[ yy_1 = 2(x + x_1). \] Similarly, for \( x^2 = -32y \), the general equation of a tangent at \( (x_2, y_2) \) is: \[ xx_2 = -16(y + y_2). \] Step 3: Find the Equation of the Common Tangent
The common tangent to both parabolas satisfies the slopes from both equations: \[ \frac{y}{2} = \frac{x}{-16}. \] Cross multiplying: \[ x - 2y + 4 = 0. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{x - 2y + 4 = 0}. \]