If a variable tangent to the curve `x^2y=c^3` makes intercepts `a , bonx-a n dy-a x e s ,` respectively, then the value of `a^2b` is `27c^3` (b) `4/(27)c^3` (c) `(27)/4c^3` (d) `4/9c^3`

Options

A. `27c^(3)`
B. `(4)/(27)c^(3)`
C. `(27)/(4)c^(3)`
D. `(4)/(9)c^(3)`

Answer

To solve the problem, we need to find the value of \( a^2b \) where \( a \) and \( b \) are the x-intercept and y-intercept of the tangent to the curve given by \( x^2y = c^3 \).

### Step 1: Find the equation of the tangent
The curve can be rewritten as:
\[ y = \frac{c^3}{x^2} \]
Let \( (x_1, y_1) \) be a point on the curve. Thus,
\[ y_1 = \frac{c^3}{x_1^2} \]
Now, we need to find the derivative \( \frac{dy}{dx} \) to get the slope of the tangent line. Differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = -\frac{2c^3}{x^3} \]

### Step 2: Write the equation of the tangent line
Using the point-slope form of the equation of a line, the equation of the tangent at the point \( (x_1, y_1) \) is:
\[ y - y_1 = \left(-\frac{2c^3}{x_1^3}\right)(x - x_1) \]
Substituting \( y_1 = \frac{c^3}{x_1^2} \):
\[ y - \frac{c^3}{x_1^2} = -\frac{2c^3}{x_1^3}(x - x_1) \]

### Step 3: Find the x-intercept (a)
To find the x-intercept \( a \), set \( y = 0 \):
\[ 0 - \frac{c^3}{x_1^2} = -\frac{2c^3}{x_1^3}(x - x_1) \]
Rearranging gives:
\[ -\frac{c^3}{x_1^2} = -\frac{2c^3}{x_1^3}(x - x_1) \]
Cancelling \( -c^3 \) from both sides:
\[ \frac{1}{x_1^2} = \frac{2}{x_1^3}(x - x_1) \]
Cross-multiplying:
\[ x - x_1 = \frac{x_1^3}{2} \]
Thus:
\[ x = x_1 + \frac{x_1^3}{2} \]
This gives us the x-intercept \( a \):
\[ a = \frac{3x_1}{2} \]

### Step 4: Find the y-intercept (b)
To find the y-intercept \( b \), set \( x = 0 \):
\[ y - \frac{c^3}{x_1^2} = -\frac{2c^3}{x_1^3}(0 - x_1) \]
This simplifies to:
\[ y - \frac{c^3}{x_1^2} = \frac{2c^3}{x_1^2} \]
Thus:
\[ y = \frac{c^3}{x_1^2} + \frac{2c^3}{x_1^2} = \frac{3c^3}{x_1^2} \]
This gives us the y-intercept \( b \):
\[ b = \frac{3c^3}{x_1^2} \]

### Step 5: Calculate \( a^2b \)
Now we can calculate \( a^2b \):
\[ a^2 = \left(\frac{3x_1}{2}\right)^2 = \frac{9x_1^2}{4} \]
Thus:
\[ a^2b = \frac{9x_1^2}{4} \cdot \frac{3c^3}{x_1^2} = \frac{27c^3}{4} \]

### Conclusion
The value of \( a^2b \) is:
\[ \frac{27c^3}{4} \]
Thus, the correct answer is (c) \( \frac{27}{4}c^3 \).