The points on the curve 9y2 = x3, 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 9y2 = x3
Differentiating with respect to x, we have:
9 (2y) dy/dx = 3x2
⇒ dy/dx = x2/6y
The slope of the normal to the given curve at point (x1, y1) is
(- 1) / dy/dx] (x1, y1) = 6y1 / x12
The equation of the normal to the curve at (x1, y1) is
y - y1 = - 6y1 (x - x1)
⇒ x12y - x12 y1 = - 6xy1 + 6x1y1
⇒ 6xy1 + x12 y = 6x1y1 + x12 y1
⇒ (6xy1 + x21y)/(6x1y1 + x12 y1) = 1
⇒ (6xy1) / (6x1y1 + x12 y1) + (x12 y) / (6x1y1 + x12 y1) = 1
⇒ x / x1(6 + x1) / 6 + y / y1 (6 + x1) / x1 = 1
It is given that the normal makes equal intercepts with the axes.
Therefore, we have:
⇒ x1 (6 + x1) / 6 = y1 (6 + x1) / x1
⇒ x1/6 = y1/x1
⇒ x21 = 6y1 ....(1)
Also, the point (x1, y1) lies on the curve, so we have
9y21 = x13 ....(2)
From (1) and (2), we have:
9 (x12 / 6) = x13
x12 / 4 = x13
x1 = 4
From (2), we have:
9 y12 = 43
y12 = 64/9
y12 = ± 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 9y2 = x3, 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 9y2 = x3, 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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