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(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

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