Question

The slope of the line touching both the parabolas $y^2 = 4x$ and $x^2 = - 32y$ is :

• A. $\frac{1}{8}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The Correct Option is C

To determine the slope of the line that touches both the parabolas \( y^2 = 4x \) and \( x^2 = -32y \), we follow these steps:

1. First, consider the parabola \( y^2 = 4x \). This is a standard form of a parabola that opens to the right. The equation of the tangent line to this parabola at any point \((x_1, y_1)\) is given by:
2. \(yy_1 = 2(x + x_1)\)
3. Next, consider the parabola \( x^2 = -32y \). This is a parabola that opens downwards. The equation of the tangent line to this parabola at any point \((x_2, y_2)\) is:
4. \(xx_2 = -16(y + y_2)\)
5. The line touches both parabolas, implying that a common tangent satisfies both tangent conditions at the points of contact. For this to be true, the slopes of these two tangents must be equal. Let the slope of the tangent be \( m \).
6. For the tangent to the first parabola, we replace \( y_1 \) with \( mx_1 \), as \( y_1 = mx_1 \). Substituting in the tangent equation:
7. \(y = mx = 2(x + x_1)\)
8. Simplifying, we have:
9. \(m^2x_1 = 2x_1\) \(\rightarrow m^2 = 2\)
10. Similarly, for the tangent to the second parabola: \( x_2 = -16(mx_2) \) \(\rightarrow m = -\frac{x_2}{16} \)
11. Equating the two expressions of \( m \) from both parabolas gives us: \(m^2 = 2\) and \(-\frac{1}{m} = 16m\)
12. Solving for \( m \), we find that \( m = \frac{1}{2} \).
13. Therefore, the slope of the line touching both parabolas is \(\frac{1}{2}\).