Question

The point on the curve \(9y^2 = x^3\) where the normal to the curve makes equal intercepts with the co-ordinate axes is

• A. \((-4,\frac{8}{3})\)
• B. \((4,\frac{8}{3})\)
• C. \((-4,\frac{-8}{3})\)
• D. \((-4,\frac{3}{8})\)

Hint:

Equal intercepts on the axes means the normal has slope -1.

Answer 1

The Correct Option is B

Step 1: Normal slope
The normal slope is $-6y/x^2$. Equal intercepts need this to equal $-1$, so $6y = x^2$.

Step 2: Intersect with the curve
$9y^2 = x^3$ with $y = x^2/6$ gives $x^4/4 = x^3$, so $x = 4$.

Step 3: Coordinates
$y = 16/6 = 8/3$. The point $(4, 8/3)$ is option (B).

Final Answer:
Option (B). \[ \boxed{\left(4,\frac{8}{3}\right)} \]