Find the point on the curve $9 y^{2} = x^{3}$ where normal to the curve has non zero x-intercept and both x-intercept and y-intercept are equal.

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Solution

$Given that slope of normal = \frac{- 1}{(\frac{d y}{d x})} = - 1$
$\frac{d y}{d x} = 1 . . . . . (1)$
$9 y^{2} = x^{3}$
$18 y \frac{d y}{d x} = 3 x^{2}$
$\frac{d y}{d x} = \frac{x^{2}}{6 y} = 1$
$x^{2} = 6 y$
$let x = a$ $t h e n$ $y = \frac{a^{2}}{6}$
$given that 9 y^{2} = x^{3}$
$9 (\frac{a^{2}}{6})^{2} = a^{3}$
$from this we get a = 4, a = 0$ $(not possible for given condition of non zero x -intercept)$
$h e n c e a = 4$
$so the point will be (4, \frac{8}{3})$

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(a)

(4, \pm \frac{8}{3})

(b)

(4, - \frac{8}{3})

(c)

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(d)

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(a) 4, $\frac{8}{3}$
(b) -4, $\frac{8}{3}$
(c) 4, $\frac{- 8}{3}$
(d) none of these

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Find the point on the curve 9y2=x3 where normal to the curve has non zero x-intercept and both x-intercept and y-intercept are equal.

Given that slope of normal =−1(dydx)=−1dydx=1.....(1)9y2=x318ydydx=3x2dydx=x26y=1x2=6ylet x=a then y=a26given that 9y2=x3 9(a26)2=a3from this we get a=4,a=0 (not possible for given condition of non zero x -intercept)hence a=4so the point will be (4,83)

Find the point on the curve $9 y^{2} = x^{3}$ where normal to the curve has non zero x-intercept and both x-intercept and y-intercept are equal.