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(D) Given curve: $27x^2 = 4y^3$.Differentiating with respect to $x$: $54x = 12y^2 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{9x}{2y^2}$.Let the p

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$x = \pm \sqrt{2}(y - 2)$

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(D) Given curve: $27x^2 = 4y^3$.Differentiating with respect to $x$: $54x = 12y^2 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{9x}{2y^2}$.Let the point on the curve be $(2t^3, 3t^2)$.Slope of the tangent at $t$: $m = \frac{9(2t^3)}{2(3t^2)^2} = \frac{18t^3}{18t^4} = \frac{1}{t}$.Equation of the tangent at $t$: $y - 3t^2 = \frac{1}{t}(x - 2t^3) \Rightarrow x - ty = -t^3$  .......$(i)$.Equation of the normal at $t_1$: $y - 3t_1^2 = -t_1(x - 2t_1^3) \Rightarrow t_1x + y = 3t_1^2 + 2t_1^4$  .......$(ii)$.Since the line is both tangent and normal,comparing $(i)$ and $(ii)$ gives $\frac{1}{t_1} = -t = \frac{-t^3}{3t_1^2 + 2t_1^4}$.From $\frac{1}{t_1} = -t$,we have $t_1 = -\frac{1}{t}$.Substituting into $-t = \frac{-t^3}{3t_1^2 + 2t_1^4}$: $t = \frac{t^3}{3(\frac{1}{t^2}) + 2(\frac{1}{t^4})} = \frac{t^7}{3t^2 + 2}$.$3t^2 + 2 = t^6 \Rightarrow t^6 - 3t^2 - 2 = 0$. Let $u = t^2$,then $u^3 - 3u - 2 = 0$.$(u - 2)(u + 1)^2 = 0$. Since $u = t^2 > 0$,we have $t^2 = 2$,so $t = \pm \sqrt{2}$.Substituting $t^2 = 2$ into the tangent equation $x = t(y - t^2)$: $x = \pm \sqrt{2}(y - 2)$.

The equations of straight lines which are both tangent and normal to the curve $27x^2 = 4y^3$ are

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