Practicing Success
If the tangent at a point P, with parameter t, on the curve $x=4 t^2+3, y=8 t^3-1, t \in R$, meets the curve again at Q, then the coordinates of Q are |
$\left(t^2+3,-t^3-1\right)$ $\left(t^2+3, t^3-1\right)$ $\left(16 t^2+3,-64 t^3-1\right)$ $\left(4 t^2+3,-8 t^3-1\right)$ |
$\left(t^2+3,-t^3-1\right)$ |
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 $\left(t^2+3,-t^3-1\right)$. |