Maximum number of common normals of `y^2=4ax and x^2=4by` is ____

Options

A. 3
B. 4
C. 6
D. 5

Explanation

To find the maximum number of common normals of the parabolas \(y^2 = 4ax\) and \(x^2 = 4by\), we can follow these steps:

### Step 1: Write the equations of the normals for both parabolas.

For the parabola \(y^2 = 4ax\), the equation of the normal in slope-intercept form is given by:
\[ y = mx - 2am - \frac{a}{m^2} \]

For the parabola \(x^2 = 4by\), the equation of the normal is:
\[ y = mx + 2b + \frac{b}{m^2} \]

### Step 2: Set the two equations equal to each other.

Since we are looking for common normals, we set the two equations equal to each other:
\[ mx - 2am - \frac{a}{m^2} = mx + 2b + \frac{b}{m^2} \]

### Step 3: Simplify the equation.

Cancel \(mx\) from both sides:
\[ -2am - \frac{a}{m^2} = 2b + \frac{b}{m^2} \]

Rearranging gives:
\[ -2am - 2b = \frac{a}{m^2} + \frac{b}{m^2} \]

Combining the terms on the right side:
\[ -2am - 2b = \frac{a + b}{m^2} \]

### Step 4: Multiply through by \(m^2\) to eliminate the fraction.

Multiplying both sides by \(m^2\) results in:
\[ -2am^3 - 2bm^2 = a + b \]

### Step 5: Rearrange to form a polynomial equation.

Rearranging gives us:
\[ -2am^3 - 2bm^2 - (a + b) = 0 \]

### Step 6: Identify the degree of the polynomial.

This is a cubic polynomial in \(m\):
\[ -2am^3 - 2bm^2 - (a + b) = 0 \]
which is of degree 3.

### Step 7: Determine the maximum number of solutions.

A cubic polynomial can have at most 3 real roots. Therefore, the maximum number of common normals is determined by the number of distinct slopes \(m\) that satisfy this equation.

### Conclusion

Thus, the maximum number of common normals to the parabolas \(y^2 = 4ax\) and \(x^2 = 4by\) is:
\[ \boxed{3} \]