# The points on the curve 9y^{2} = x^{3}, where the normal to the curve makes equal intercepts with the axes are

(A) (4, ± 8/3) (B) (4, - 8/3) (C) (4, ± 3/8) (D) (± 4, 3/8)

**Solution:**

The equation of the given curve is 9y^{2} = x^{3}

Differentiating with respect to x, we have:

9 (2y) dy/dx = 3x^{2}

⇒ dy/dx = x^{2}/6y

The slope of the normal to the given curve at point (x_{1}, y_{1}) is

(- 1) / dy/dx] _{(x1, y1)} = 6y_{1 }/ x_{1}^{2}

The equation of the normal to the curve at (x_{1}, y_{1}) is

y - y_{1} = - 6y_{1} (x - x_{1})

⇒ x_{1}^{2}y - x_{1}^{2} y_{1} = - 6xy_{1} + 6x_{1}y_{1}

⇒ 6xy_{1} + x_{1}^{2 }y = 6x_{1}y_{1} + x_{1}^{2} y_{1}

⇒ (6xy_{1} + x^{2}_{1}y)/(6x_{1}y_{1} + x_{1}^{2} y_{1}) = 1

⇒ (6xy_{1}) / (6x_{1}y_{1} + x_{1}^{2 } y_{1}) + (x_{1}^{2 }y) / (6x_{1}y_{1} + x_{1}^{2} y_{1}) = 1

⇒ x / x_{1}(6 + x_{1}) / 6 + y / y_{1 }(6 + x_{1}) / x_{1} = 1

It is given that the normal makes equal intercepts with the axes.

Therefore, we have:

⇒ x_{1 }(6 + x_{1}) / 6 = y_{1 }(6 + x_{1}) / x_{1}

⇒ x_{1}/6 = y_{1}/x_{1}

⇒ x^{2}_{1} = 6y_{1} ....(1)

Also, the point (x_{1}, y_{1}) lies on the curve, so we have

9y^{2}_{1} = x_{1}^{3} ....(2)

From (1) and (2), we have:

9 (x_{1}^{2} / 6) = x_{1}^{3}

x_{1}^{2} / 4 = x_{1}^{3}

x_{1} = 4

From (2), we have:

9 y_{1}^{2} = 4^{3}

y_{1}^{2} = 64/9

y_{1}^{2} = ± 8 / 3

Hence, the required points are (4, ± 8/3)

Thus, the correct option is A

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 24

## The points on the curve 9y^{2} = x^{3}, where the normal to the curve makes equal intercepts with the axes are (A) (4, ± 8/3) (B) (4, - 8/3) (C) (4, ± 3/8) (D) (± 4, 3/8)

**Summary:**

The points on the curve 9y^{2} = x^{3}, where the normal to the curve makes equal intercepts with the axes are (4, ± 8/3). Thus, the correct option is A

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