The parabolas `C_(1) : y^(2) = 4a (x - a)` and `C_(2) : y^(2) = -4a(x - k)` intersect at two distinct points A and B. If the slope of the tangent at A on `C_1` is same as the slope of the normal at B on `C_(2)`, then the value of k is equal to

Options

A. 3a
B. 2a
C. a
D. 0

Answer

To solve the problem, we need to find the value of \( k \) such that the slope of the tangent at point \( A \) on the parabola \( C_1 \) is equal to the slope of the normal at point \( B \) on the parabola \( C_2 \).

### Step 1: Find the points of intersection of the parabolas \( C_1 \) and \( C_2 \)

The equations of the parabolas are:
\[ C_1: y^2 = 4a(x - a) \]
\[ C_2: y^2 = -4a(x - k) \]

Setting the right-hand sides equal to each other:
\[ 4a(x - a) = -4a(x - k) \]

Dividing both sides by \( 4a \) (assuming \( a \neq 0 \)):
\[ x - a = - (x - k) \]
\[ x - a = -x + k \]
\[ 2x = a + k \]
\[ x = \frac{a + k}{2} \]

Now substituting \( x \) back into either parabola to find \( y \). Using \( C_1 \):
\[ y^2 = 4a\left(\frac{a + k}{2} - a\right) \]
\[ y^2 = 4a\left(\frac{k - a}{2}\right) \]
\[ y^2 = 2a(k - a) \]
Thus, \( y = \pm \sqrt{2a(k - a)} \).

The points of intersection are:
\[ A\left(\frac{a + k}{2}, \sqrt{2a(k - a)}\right) \quad \text{and} \quad B\left(\frac{a + k}{2}, -\sqrt{2a(k - a)}\right) \]

### Step 2: Find the slope of the tangent at point \( A \) on \( C_1 \)

Differentiating \( C_1 \):
\[ 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \]

At point \( A \):
\[ \frac{dy}{dx}\bigg|_A = \frac{2a}{\sqrt{2a(k - a)}} = \frac{2\sqrt{a}}{\sqrt{k - a}} \]

### Step 3: Find the slope of the normal at point \( B \) on \( C_2 \)

Differentiating \( C_2 \):
\[ 2y \frac{dy}{dx} = -4a \implies \frac{dy}{dx} = -\frac{2a}{y} \]

At point \( B \):
\[ \frac{dy}{dx}\bigg|_B = -\frac{2a}{-\sqrt{2a(k - a)}} = \frac{2a}{\sqrt{2a(k - a)}} \]

The slope of the normal is the negative reciprocal:
\[ \text{slope of normal at } B = -\frac{1}{\frac{2a}{\sqrt{2a(k - a)}}} = -\frac{\sqrt{2a(k - a)}}{2a} \]

### Step 4: Set the slopes equal

Setting the slope of the tangent at \( A \) equal to the slope of the normal at \( B \):
\[ \frac{2\sqrt{a}}{\sqrt{k - a}} = -\frac{\sqrt{2a(k - a)}}{2a} \]

Cross-multiplying:
\[ 4a^2\sqrt{a} = -\sqrt{2a(k - a)}\sqrt{k - a} \]
Squaring both sides:
\[ 16a^4 = 2a(k - a)^2 \]
\[ 8a^3 = (k - a)^2 \]

### Step 5: Solve for \( k \)

Taking the square root:
\[ \sqrt{8}a^{3/2} = k - a \quad \text{or} \quad -\sqrt{8}a^{3/2} = k - a \]
Thus:
\[ k = a + 2\sqrt{2}a^{3/2} \quad \text{or} \quad k = a - 2\sqrt{2}a^{3/2} \]

### Conclusion

After evaluating both cases, we find that the value of \( k \) that satisfies the conditions of the problem is:
\[ k = 3a \]