Question

The maximum number of common normals of \( y^2 = 4ax \) and \( x^2 = 4by \) is equal to:

• A. \( 3 \)
• B. \( 4 \)
• C. \( 5 \)
• D. \( 6 \)

Hint:

When finding common normals to two conic sections, solving for the slopes of the normals from both curves typically gives the number of common normals. The number of solutions depends on the geometry and relative positioning of the curves.

Answer 1

The Correct Option is C

We are given the equations of two parabolas:
\( y^2 = 4ax \), a rightward-opening parabola.
\( x^2 = 4by \), an upward-opening parabola.
We need to determine the maximum number of common normals to both curves.
Step 1: General equation of the normal to the parabola \( y^2 = 4ax \). For the parabola \( y^2 = 4ax \), the general equation of the normal at a point \( (x_1, y_1) \) on the parabola is: \[ y - y_1 = -\frac{x_1}{y_1} (x - x_1). \] This normal has slope \( -\frac{x_1}{y_1} \).
Step 2: General equation of the normal to the parabola \( x^2 = 4by \).
For the parabola \( x^2 = 4by \), the equation of the normal at a point \( (x_2, y_2) \) on this parabola is: \[ y - y_2 = \frac{y_2}{x_2} (x - x_2). \] This normal has slope \( \frac{y_2}{x_2} \).
Step 3: Condition for common normals.
For the normals to be common to both parabolas, their slopes must be equal. Thus, the condition is: \[ -\frac{x_1}{y_1} = \frac{y_2}{x_2}. \] This leads to a system of equations that can give multiple solutions. The maximum number of solutions, considering the different geometric configurations and the relative positions of the two parabolas, is 5.
Step 4: Conclusion.
The maximum number of common normals to the given parabolas is 5.