If the tangent at the point P with coordinate | vedclass

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(B) Given the curve $y^{2}=2x^{3}$. Differentiating with respect to $x$, we get $2y \frac{dy}{dx} = 6x^{2}$, which implies $\frac{dy}{dx} = \frac{3x^{

Vedclass · Accepted Answer

$(h, k)=\left(\frac{1}{8},-\frac{1}{16}\right)$ only

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(B) Given the curve $y^{2}=2x^{3}$. Differentiating with respect to $x$, we get $2y \frac{dy}{dx} = 6x^{2}$, which implies $\frac{dy}{dx} = \frac{3x^{2}}{y}$.At point $P(h, k)$, the slope of the tangent is $m_{1} = \frac{3h^{2}}{k}$.The given line is $4x=3y$, or $y=\frac{4}{3}x$, which has a slope $m_{2} = \frac{4}{3}$.Since the tangent is perpendicular to the line, $m_{1} 	imes m_{2} = -1$.So, $\left(\frac{3h^{2}}{k}ight) 	imes \left(\frac{4}{3}ight) = -1 \Rightarrow \frac{4h^{2}}{k} = -1 \Rightarrow k = -4h^{2}$.Since $P(h, k)$ lies on the curve, $k^{2} = 2h^{3}$.Substituting $k = -4h^{2}$ into the curve equation: $(-4h^{2})^{2} = 2h^{3} \Rightarrow 16h^{4} = 2h^{3}$.This gives $2h^{3}(8h - 1) = 0$, so $h=0$ or $h=\frac{1}{8}$.If $h=0$, then $k=0$. However, at $(0,0)$, the derivative $\frac{dy}{dx}$ is undefined (the curve has a cusp), so the tangent is not well-defined in the standard sense.If $h=\frac{1}{8}$, then $k = -4(\frac{1}{8})^{2} = -4(\frac{1}{64}) = -\frac{1}{16}$.Thus, the point is $\left(\frac{1}{8}, -\frac{1}{16}ight)$.

If the tangent at the point $P$ with coordinates $(h, k)$ on the curve $y^{2}=2x^{3}$ is perpendicular to the straight line $4x=3y$, then

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