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(D) The coordinates of point $P$ are $(4t^2 + 3, 8t^3 - 1).$ The slope of the tangent at $P$ is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{

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$(t^2 + 3, -t^3 - 1)$

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(D) The coordinates of point $P$ are $(4t^2 + 3, 8t^3 - 1).$ The slope of the tangent at $P$ is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{24t^2}{8t} = 3t.$ Let the coordinates of point $Q$ be $(4\lambda^2 + 3, 8\lambda^3 - 1).$ The slope of the line $PQ$ is $\frac{8\lambda^3 - 8t^3}{4\lambda^2 - 4t^2} = \frac{2(\lambda^3 - t^3)}{\lambda^2 - t^2} = \frac{2(\lambda - t)(\lambda^2 + \lambda t + t^2)}{(\lambda - t)(\lambda + t)} = \frac{2(\lambda^2 + \lambda t + t^2)}{\lambda + t}.$ Since $PQ$ is the tangent at $P,$ its slope must be $3t.$ So,$\frac{2(\lambda^2 + \lambda t + t^2)}{\lambda + t} = 3t.$ $2\lambda^2 + 2\lambda t + 2t^2 = 3t\lambda + 3t^2.$ $2\lambda^2 - t\lambda - t^2 = 0.$ $(2\lambda + t)(\lambda - t) = 0.$ Since $\lambda 
eq t$ (as $Q$ is a different point),we have $\lambda = -t/2.$ Substituting $\lambda = -t/2$ into the coordinates of $Q,$ we get $x = 4(-t/2)^2 + 3 = t^2 + 3$ and $y = 8(-t/2)^3 - 1 = -t^3 - 1.$ Thus,the coordinates of $Q$ are $(t^2 + 3, -t^3 - 1).$

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

If the tangent at a point $P,$ with parameter $t,$ on the curve $x = 4t^2 + 3, y = 8t^3 - 1, t \in R,$ meets the curve again at a point $Q,$ then the coordinates of $Q$ are

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