The slope of normal to the curve ` x^(3)=8a^(2)y, a gt 0` at a point in the first quadrant is `-(2)/(3)` , then point is

Options

A. (2a,-a)
B. (2a,a)
C. (a,2a)
D. (a,a)

Answer

To solve the problem, we need to find the point on the curve \( x^3 = 8a^2 y \) where the slope of the normal is \( -\frac{2}{3} \).

### Step 1: Differentiate the curve
Given the equation of the curve:
\[ x^3 = 8a^2 y \]
we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^3) = \frac{d}{dx}(8a^2 y) \]
This gives us:
\[ 3x^2 = 8a^2 \frac{dy}{dx} \]
Let \( \frac{dy}{dx} = y' \). Therefore, we can express \( y' \) as:
\[ y' = \frac{3x^2}{8a^2} \]

### Step 2: Find the slope of the normal
The slope of the normal \( m_2 \) is related to the slope of the tangent \( m_1 \) by the formula:
\[ m_2 = -\frac{1}{m_1} \]
Substituting \( m_1 = y' \):
\[ m_2 = -\frac{1}{y'} = -\frac{8a^2}{3x^2} \]

### Step 3: Set the slope of the normal equal to the given slope
We know from the problem that the slope of the normal is \( -\frac{2}{3} \):
\[ -\frac{8a^2}{3x^2} = -\frac{2}{3} \]
Removing the negative signs and simplifying:
\[ \frac{8a^2}{3x^2} = \frac{2}{3} \]

### Step 4: Cross-multiply and solve for \( x^2 \)
Cross-multiplying gives:
\[ 8a^2 = 2x^2 \]
Dividing both sides by 2:
\[ 4a^2 = x^2 \]
Thus, we find:
\[ x = \pm 2a \]
Since we are looking for a point in the first quadrant, we take:
\[ x = 2a \]

### Step 5: Find the corresponding \( y \) value
Substituting \( x = 2a \) back into the original curve equation to find \( y \):
\[ (2a)^3 = 8a^2 y \]
This simplifies to:
\[ 8a^3 = 8a^2 y \]
Dividing both sides by \( 8a^2 \) (since \( a > 0 \)):
\[ y = a \]

### Step 6: State the point
Thus, the point on the curve in the first quadrant is:
\[ (2a, a) \]

### Final Answer
The point is \( (2a, a) \).