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: Understanding the Concept
A line making equal intercepts with the coordinate axes has slope \(-1\) (it is of the form \(x + y = k\)). So the normal to the curve must have slope \(-1\), which means the tangent has slope 1.

Step 2: Slope of the tangent
From \(9y^2 = x^3\): \(18y\,\dfrac{dy}{dx} = 3x^2\), so \(\dfrac{dy}{dx} = \dfrac{x^2}{6y}\).
Set it equal to 1: \(x^2 = 6y\), so \(y = \dfrac{x^2}{6}\).

Step 3: Locate the point
Substitute in the curve:
\[ 9\cdot\frac{x^4}{36} = x^3 \Rightarrow \frac{x^4}{4} = x^3 \Rightarrow x = 4\ (x \neq 0) \]
\[ y = \frac{16}{6} = \frac83 \]
The point is \(\left(4, \frac83\right)\), option (B). Check: \(9\cdot\frac{64}{9} = 64 = 4^3\). Options with \(x = -4\) give \(x^3 = -64 < 0\) while \(9y^2 \geq 0\), so they are not on the curve.

Final Answer:
The point is \(\left(4, \frac83\right)\), option (B). \[ \boxed{\left(4,\frac{8}{3}\right)} \]