If the tangent to the curve $\mathrm{y}=\mathrm{x}^{3}$ at the point $\mathrm{P}\left(\mathrm{t}, \mathrm{t}^{3}\right)$ meets the curve again at $\mathrm{Q}$, then the ordinate of the point which divides PQ internally in the ratio $1: 2$ is:
Correct Option: 1
Slope of tangent at $\left.P\left(t, t^{3}\right)=\frac{d y}{d x}\right]_{\left(t, t^{3}\right)}$
$=\left(3 x^{2}\right)_{x=t}=3 t^{2}$
So equation tangent at $\mathrm{P}\left(\mathrm{t}, \mathrm{t}^{3}\right)$ :
$y-t^{3}=3 t^{2}(x-t)$
for point of intersection with $\mathrm{y}=\mathrm{x}^{3}$
$x^{3}-t^{3}=3 t^{2} x-3 t^{3}$
$\Rightarrow \quad(x-t)\left(x^{2}+x t+t^{2}\right)=3 t^{2}(x-t)$
for $x \neq t$
$x^{2}+x t+t^{2}=3 t^{2}$
$\Rightarrow x^{2}+x t-2 t^{2}=0 \Rightarrow(x-t)(x+2 t)=0$
So for $Q: x=-2 t, Q\left(-2 t,-8 t^{3}\right)$
ordinate of required point $: \frac{2 t^{3}-8 t^{3}}{2+1}=-2 t^{3}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.