At which point on the curve 9y^2 = x^3 does t | vedclass

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(A) Let the required point be $(x_1, y_1)$.The equation of the curve is $9y^2 = x^3$.Since $(x_1, y_1)$ lies on the curve,we have $9y_1^2 = x_1^3$ ...

Vedclass · Accepted Answer

$4, \pm \frac{8}{3}$

Vedclass · Answer

(A) Let the required point be $(x_1, y_1)$.The equation of the curve is $9y^2 = x^3$.Since $(x_1, y_1)$ lies on the curve,we have $9y_1^2 = x_1^3$ ... $(i)$.Differentiating $9y^2 = x^3$ with respect to $x$,we get $18y \frac{dy}{dx} = 3x^2$,which implies $\frac{dy}{dx} = \frac{x^2}{6y}$.The slope of the tangent at $(x_1, y_1)$ is $m_t = \frac{x_1^2}{6y_1}$.The slope of the normal at $(x_1, y_1)$ 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 $\pm 1$.Therefore,$-\frac{6y_1}{x_1^2} = \pm 1$,which implies $x_1^2 = \mp 6y_1$.Squaring both sides,$x_1^4 = 36y_1^2$. Substituting $y_1^2 = \frac{x_1^3}{9}$ from $(i)$,we get $x_1^4 = 36 \left( \frac{x_1^3}{9} \right) = 4x_1^3$.This gives $x_1^3(x_1 - 4) = 0$,so $x_1 = 0$ or $x_1 = 4$.If $x_1 = 0$,then $y_1 = 0$. However,the normal at $(0,0)$ is not well-defined in a way that creates equal intercepts (it would pass through the origin).If $x_1 = 4$,then $9y_1^2 = 4^3 = 64$,so $y_1^2 = \frac{64}{9}$,which gives $y_1 = \pm \frac{8}{3}$.Thus,the required points are $(4, 8/3)$ and $(4, -8/3)$.

At which point on the curve $9y^2 = x^3$ does the normal to the curve make equal intercepts with the axes?

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