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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 a point Q, then the coordinates of Q are :
(a) $({t^2} + 3, - {t^2} - 1)$
(b) $(4{t^2} + 3, - 8{t^3} - 3)$
(c) $(16{t^3} + 3, - 64{t^3} - 1)$
(d) $({t^2} + 3,{t^3} - 1)$

Answer Verified Verified
Hint:In this question first try to find out the equation of tangent in the term of parameter t then take a point Q as ($4{t_1}^2 + 3$,$8{t_1}^3 - 1$) now put this point on the equation of tangent on solving find out the relation between ${t_1}$ and $t$ then put the value in point Q.

Complete step-by-step answer:
It is given that the x = $4{t^2} + 3$ , y = $8{t^3} - 1$ , t$ \in $ R
Differentiate x and y with respect to t .
$\dfrac{{dx}}{{dt}} = 8t$ and $\dfrac{{dy}}{{dt}} = 24{t^2}$
Hence the slope of the curve at point P is
$\dfrac{{dy}}{{dx}} = 3t$
Equation of tangent = $(y - {y_1}) = \dfrac{{dy}}{{dx}}(x - {x_1})$
At point P ($4{t^2} + 3$,$8{t^3} - 1$)
Equation of tangent become = $(y - 8{t^3} + 1) = t(x - 4{t^2} - 3)$
Hence this tangent meets the curve once again at point Q as ($4{t_1}^2 + 3$,$8{t_1}^3 - 1$) .
By putting the coordinate in the equation of tangent .
\[(8{t_1}^3 - 1 - 8{t^3} + 1) = 3t(4{t_1}^2 + 3 - 4{t^2} - 3)\]
\[8({t_1}^3 - {t^3}) = 24t({t_1}^2 - {t^2})\]
\[2({t_1}^3 - {t^3}) = 3t({t_1}^2 - {t^2})\]
After expanding the equation as ${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$ we get
$2({t_1} - t)({t_1}^2 + {t^2} + t{t_1}) = 3t({t_1} - t)({t_1} + t)$
As ${t_1} \ne t$ because it represent same point that is P
$2({t_1}^2 + {t^2} + t{t_1}) = 3t({t_1} + t)$
$2{t_1}^2 + 2{t^2} + 2t{t_1} = 3t{t_1} + {t^2}$
$2{t_1}^2 + {t^2} + t{t_1} = 0$
Separate the ${t_1}$
${t_1}^2 + {t^2} + {t_1}^2 + t{t_1} = 0$
$({t_1} - t)({t_1} + t) + {t_1}({t_1} - t) = 0$
$2{t_1} + t = 0$
${t_1} = - \dfrac{t}{2}$
After putting the value of ${t_1} = - \dfrac{t}{2}$ in the point Q as ($4{t_1}^2 + 3$,$8{t_1}^3 - 1$)
After assign the value we get Q $({t^2} + 3, - {t^2} - 1)$

So, the correct answer is “Option A”.

Note:Always remember that the slope of tangent is $\dfrac{{dy}}{{dx}}$, if in the question x and y terms are given in the form of any other variable then differentiate with respect to that term and finally divide to get $\dfrac{{dy}}{{dx}}$.In some questions the tangents intersect two points on the curve then proceed the same as the given question and take third point as parametric coordinate ${t_3}$ and find relation in between .