Point on the curve $9 y^{2} = x^{3}$ , where the normal to the curve makes equal intercepts on co-ordinate axes, is

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Solution

REF. Image.
Solution -
$9 y^{2} = x^{3}$
$y = \frac{x^{3 / 2}}{3}$
$\frac{d y}{d x} = \frac{3}{2} \times \frac{x^{1 / 2}}{3} = \frac{1}{2} \sqrt{x} .$
slope of Normal = $\frac{- 2}{\sqrt{x}} = - 1$
$x = 4.$
$y = \sqrt{\frac{64}{9}} = \frac{8}{3}$
point is (4, 8/3)

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9 y^{2} = x^{3}

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The point on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes is
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(b) -4, $\frac{8}{3}$
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Q. The points on the curve 9y² = x³, where the normal to the curve make equal intercepts with the axes are

(a)

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

(b)

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(c)

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(d)

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Point on the curve 9y2=x3, where the normal to the curve makes equal intercepts on co-ordinate axes, is

REF. Image.Solution - 9y2=x3y=x3/23dydx=32×x1/23=12√x.slope of Normal = −2√x=−1x=4.y=√649=83point is (4, 8/3)

Point on the curve $9 y^{2} = x^{3}$ , where the normal to the curve makes equal intercepts on co-ordinate axes, is

REF. Image.
Solution -
$9 y^{2} = x^{3}$
$y = \frac{x^{3 / 2}}{3}$
$\frac{d y}{d x} = \frac{3}{2} \times \frac{x^{1 / 2}}{3} = \frac{1}{2} \sqrt{x} .$
slope of Normal = $\frac{- 2}{\sqrt{x}} = - 1$
$x = 4.$
$y = \sqrt{\frac{64}{9}} = \frac{8}{3}$
point is (4, 8/3)