Question

The equation of the common tangent to the parabolas \(y^{2}=4x\) and \(x^{2}=32y\) is ...

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

Hint:

Write the tangent to the first parabola with slope m and apply the tangency condition to the second.

Answer 1

The Correct Option is C

Step 1: Reverse approach
For $x^{2}=32y$, the tangent is $y = mx - 8m^{2}$. For $y^{2}=4x$ the intercept must be $c = \dfrac{1}{m}$.

Step 2: Solve
$-8m^{2}=\dfrac{1}{m}$ gives $m^{3}=-\dfrac{1}{8}$, so $m=-\dfrac{1}{2}$ and $c=-2$.

Step 3: Result
The tangent is $y=-\dfrac{x}{2}-2$, i.e. $x+2y+4=0$.

Final Answer:
The common tangent line is x + 2y + 4 = 0. \[ \boxed{\text{(C) }x+2y+4=0} \]