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. `x^(2)-y^(2)=a^(2)`
B. `xy=a^(2)`
C. `x^(2)+y^(2)=a^(2)`
D. `xy=4a^(2)`

Answer

To solve the problem, we need to find the locus of the point of contact of the two parabolas given by the equations \( y^2 = 4a(x - \lambda_1) \) and \( x^2 = 4a(y - \lambda_2) \).

### Step-by-step Solution:

**Step 1: Identify 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 parabola \( y^2 = 4a(x - \lambda_1) \):
- Differentiate both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 4a \]
- Thus, the slope \( m_1 \) of the tangent at point \( (h, k) \) is:
\[ m_1 = \frac{4a}{2k} = \frac{2a}{k} \]

**Hint for Step 2:** Remember that the slope of the tangent line can be found by implicit differentiation.

**Step 3: Find the slope of the tangent to the second parabola.**
For the parabola \( x^2 = 4a(y - \lambda_2) \):
- Differentiate both sides with respect to \( y \):
\[ 2x \frac{dx}{dy} = 4a \]
- Thus, the slope \( m_2 \) of the tangent at point \( (h, k) \) is:
\[ m_2 = \frac{4a}{2h} = \frac{2a}{h} \]

**Hint for Step 3:** Use implicit differentiation to find the slope of the tangent for the second parabola.

**Step 4: Set the slopes equal for the common tangent.**
Since the parabolas touch each other, the slopes of their tangents at the point of contact must be equal:
\[ \frac{2a}{k} = \frac{2a}{h} \]
- Simplifying gives:
\[ h = k \]

**Hint for Step 4:** When two curves touch, their tangents at the point of contact are equal, leading to a relationship between \( h \) and \( k \).

**Step 5: Substitute \( k \) in terms of \( h \) into one of the parabola equations.**
Using the equation of the first parabola:
\[ k^2 = 4a(h - \lambda_1) \]
Substituting \( k = h \):
\[ h^2 = 4a(h - \lambda_1) \]
Rearranging gives:
\[ h^2 = 4ah - 4a\lambda_1 \]
or
\[ h^2 - 4ah + 4a\lambda_1 = 0 \]

**Hint for Step 5:** Rearranging the equation helps in finding a relationship that can be used to eliminate \( \lambda_1 \).

**Step 6: Similarly, substitute \( h \) in terms of \( k \) into the second parabola equation.**
Using the equation of the second parabola:
\[ h^2 = 4a(k - \lambda_2) \]
Substituting \( k = h \):
\[ h^2 = 4a(h - \lambda_2) \]
Rearranging gives:
\[ h^2 - 4ah + 4a\lambda_2 = 0 \]

**Hint for Step 6:** This will give you a similar quadratic equation in terms of \( \lambda_2 \).

**Step 7: Eliminate the parameters \( \lambda_1 \) and \( \lambda_2 \).**
From the two equations derived, we can set them equal to each other to eliminate \( \lambda_1 \) and \( \lambda_2 \):
\[ 4a\lambda_1 = 4ah - h^2 \]
\[ 4a\lambda_2 = 4ah - h^2 \]
Thus, we can conclude that both equations lead to the same relationship.

**Step 8: Find the locus.**
From the relationship \( hk = 4a^2 \) (derived from the equality of the slopes), we replace \( h \) with \( x \) and \( k \) with \( y \):
\[ xy = 4a^2 \]

**Final Result:**
The locus of the point of contact of the two parabolas is given by:
\[ xy = 4a^2 \]

**Hint for Step 8:** The locus is a hyperbola, and the relationship between \( x \) and \( y \) gives the equation of the hyperbola.