The coordinates of the point on the curve 9y^ | vedclass

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(A) Given the curve $9y^2 = x^3$. Differentiating with respect to $x$,we get $18y \frac{dy}{dx} = 3x^2$,so $\frac{dy}{dx} = \frac{x^2}{6y}$.At a point

Vedclass · Accepted Answer

$\left( 4, \frac{8}{3} \right)$

Vedclass · Answer

(A) Given the curve $9y^2 = x^3$. Differentiating with respect to $x$,we get $18y \frac{dy}{dx} = 3x^2$,so $\frac{dy}{dx} = \frac{x^2}{6y}$.At a point $(x_1, y_1)$,the slope of the tangent is $m_T = \frac{x_1^2}{6y_1}$.The slope of the normal is $m_N = -\frac{1}{m_T} = -\frac{6y_1}{x_1^2}$.Since the normal makes equal intercepts with the axes,its slope must be $\pm 1$. Given the geometry,the slope of the normal is $-1$.So,$-\frac{6y_1}{x_1^2} = -1 \implies x_1^2 = 6y_1$.Substitute this into the curve equation $9y_1^2 = x_1^3$. Since $x_1^2 = 6y_1$,we have $x_1^2 = 6(\frac{x_1^3}{9}) = \frac{2}{3}x_1^3$.For $x_1 
eq 0$,$1 = \frac{2}{3}x_1 \implies x_1 = \frac{3}{2}$. However,checking the slope condition again,if the normal makes equal intercepts,its equation is $x + y = c$ or $x - y = c$. The slope is $-1$.Solving $x_1^2 = 6y_1$ and $9y_1^2 = x_1^3$ simultaneously: $9(\frac{x_1^2}{6})^2 = x_1^3 \implies 9(\frac{x_1^4}{36}) = x_1^3 \implies \frac{x_1^4}{4} = x_1^3 \implies x_1 = 4$.Then $y_1^2 = \frac{4^3}{9} = \frac{64}{9} \implies y_1 = \pm \frac{8}{3}$.Thus,the points are $(4, \frac{8}{3})$ and $(4, -\frac{8}{3})$. Option $A$ matches.

The coordinates of the point on the curve $9y^2 = x^3$ where the normal to the curve makes equal intercepts with the axes are:

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