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The point on the curve 9y^2=x^3 , where ...

The point on the curve `9y^2=x^3` , where the normal to the curve makes equal intercepts with the axes is `(4,\ +-8//3)` (b) `(-4,\ 8//3)` (c) `(-4,\ -8//3)` (d) `(8//3,\ 4)`

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Given,`9y^2=x^3`
`y=x^(3/2)/3`
​ `dy/dx​=(3/2​)×(x^(1/2))/3​=1/2​sqrt(x) `
​. slope of Normal = `-2/sqrt(x)=-1`
​ x=4.
`y=8/3`
​ ...
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