Two parabolas `y^2 = 4a(x-lamda)` and `x^2 = 4a(y -mu)` always touch each other, then the point of contact lies on (`lamda,mu` being parameters).

Options

A. straight line
B. circle
C. parabola
D. hyperbola

Answer

To solve the problem of finding the point of contact of the two parabolas \(y^2 = 4a(x - \lambda)\) and \(x^2 = 4a(y - \mu)\) that always touch each other, we will follow these steps:

### Step 1: Differentiate both equations
We start with the two equations:
1. \(y^2 = 4a(x - \lambda)\)
2. \(x^2 = 4a(y - \mu)\)

We differentiate both equations with respect to \(x\).

For the first equation:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4a(x - \lambda)) \]
Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = 4a \cdot 1 \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \quad \text{(1)} \]

For the second equation:
\[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4a(y - \mu)) \]
Using the chain rule on the right side:
\[ 2x = 4a \frac{dy}{dx} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{2x}{4a} = \frac{x}{2a} \quad \text{(2)} \]

### Step 2: Set the derivatives equal
Since the parabolas touch each other, their slopes at the point of contact must be equal:
\[ \frac{2a}{y} = \frac{x}{2a} \]

### Step 3: Cross-multiply to find a relationship
Cross-multiplying gives:
\[ 2a \cdot 2a = xy \]
This simplifies to:
\[ 4a^2 = xy \quad \text{(3)} \]

### Step 4: Conclusion
The equation \(xy = 4a^2\) represents the condition for the point of contact of the two parabolas. Hence, the point of contact lies on the hyperbola defined by this equation.

### Final Answer
The point of contact lies on the hyperbola defined by the equation \(xy = 4a^2\).

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