Question

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)

Answer 1

Given equation is 9y² = x³

We know that Normal makes equal intercept

Given curve is 9y²=x³
Differentiating we get,

Squaring x⁴ = 36y²

Therefore, points are (4, 8/3) and (4, -8/3)

So, the option (A) (4, ±8/3) is correct

Answer 2

The equation of the given curve is 9y² = x³.
Differentiating with respect to x, we have:

The slope of the normal to the given curve at point is (x₁, y₁) is

It is given that the normal makes equal intercepts with the axes.
Therefore, We have:

The correct answer is A.

Comments

For (0,0), x-intercept = y-intercept=0, won't we include (0,0) in the answer?

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In this solution, you cancelled (6+x1) on both sides of the equation... x1 may be equal to - 6...

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x1 can not be equal to -6, see my solution below

Answer 3

Given: 9y² = x³ 

Diff. w.r.t (x) we get

Slope of normal to curve at (x₁, y₁) = -6y₁/x₁²

Normal to the curve makes equal intercepts on coordinate axes its slope = ±1

(x₁, y₁) lies on curve (1) 9y₁² = x₁³

When x₁ = 0, y₁ = 0 

when x₁ = 4 y₁ = ±8/3 as normal makes equal intercept it cannot passes through axsis.