Two parabolas `y^(2)=4a(x-lamda_(1))andx^(2)=4a(y-lamda_(2))` always touch each other (`lamda_(1),lamda_(2)` being variable parameters). Then their point of contact lies on a

Options

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

Answer

To solve the problem of finding the locus of the point of contact between the two parabolas \(y^2 = 4a(x - \lambda_1)\) and \(x^2 = 4a(y - \lambda_2)\), we will follow these steps:

### Step 1: Define the Point of Contact
Let the point of contact be \((h, k)\).

### Step 2: Find the Slope of the Tangent to the First Parabola
For the first parabola \(y^2 = 4a(x - \lambda_1)\):
1. Differentiate implicitly with respect to \(x\):
\[ 2y \frac{dy}{dx} = 4a \]
2. Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]
3. At the point of contact \((h, k)\), this becomes:
\[ \frac{dy}{dx} = \frac{2a}{k} \]

### Step 3: Find the Slope of the Tangent to the Second Parabola
For the second parabola \(x^2 = 4a(y - \lambda_2)\):
1. Differentiate implicitly with respect to \(y\):
\[ 2x = 4a \frac{dy}{dx} \]
2. Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x}{2a} \]
3. At the point of contact \((h, k)\), this becomes:
\[ \frac{dy}{dx} = \frac{h}{2a} \]

### Step 4: Set the Slopes Equal
Since the two parabolas touch each other at the point of contact, their slopes must be equal:
\[ \frac{2a}{k} = \frac{h}{2a} \]

### Step 5: Cross Multiply and Rearrange
Cross multiplying gives:
\[ 4a^2 = hk \]

### Step 6: Express the Locus
This equation can be rearranged to express the relationship between \(h\) and \(k\):
\[ hk = 4a^2 \]
Replacing \(h\) with \(x\) and \(k\) with \(y\) (since \(h\) and \(k\) represent coordinates), we have:
\[ xy = 4a^2 \]

### Step 7: Identify the Type of Curve
The equation \(xy = 4a^2\) represents a hyperbola.

### Conclusion
Thus, the locus of the point of contact between the two parabolas is a hyperbola given by the equation \(xy = C\) where \(C = 4a^2\).