If the tangent at a point P, with parameter t, on the curve x=4 t^2+3…

Let $P\left(4 t^2+3,8 t^3-1\right)$ be a point on the given curve.

Now, $x=4 t^2+3, y=8 t^3-1$

$\Rightarrow \frac{d x}{d t}=8 t$ and $\frac{d y}{d t}=24 t^2$

∴  $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{24 t^2}{8 t}=3 t$

The tangent to the curve at P is

$y-\left(8 t^3-1\right)=3 t\left(x-4 t^2-3\right)$

This will pass through $Q\left(4 t_1^2+3,8 t_1^3-1\right)$, if

$\left(8 t_1^3-1\right)-\left(8 t^3-1\right)=3 t\left(4 t_1^2+3-4 t^2-3\right)$

$\Rightarrow 8\left(t_1^3-t^3\right)=12 t\left(t_1^2-t^2\right)$

$\Rightarrow 2\left(t_1^2+t t_1+t^2\right)=3 t_0\left(t_1+t\right)$

$\Rightarrow 2 t_1^2-t_1 t-2 t^2=0$

$\Rightarrow \left(t_1-t\right)\left(2 t_1+t\right)=0$

$\Rightarrow t_1=-\frac{t}{2}$                   [∵ t₁ ≠ t]

So, the coordinates of Q are $\left(t^2+3,-t^3-1\right)$.