Solveeit Logo
Login

Question

Question: If the tangent at a point P, with parameter t, on the curve x = \(4{t^2} + 3\) , y = \(8{t^3} - 1\) ...

If the tangent at a point P, with parameter t, on the curve x = 4t2+34{t^2} + 3 , y = 8t3−18{t^3} - 1 , t∈\in R meets the curve again at a point Q, then the coordinates of Q are :
(a) (t2+3,−t2−1)({t^2} + 3, - {t^2} - 1)
(b) (4t2+3,−8t3−3)(4{t^2} + 3, - 8{t^3} - 3)
(c) (16t3+3,−64t3−1)(16{t^3} + 3, - 64{t^3} - 1)
(d) (t2+3,t3−1)({t^2} + 3,{t^3} - 1)

Explanation

Solution

In this question first try to find out the equation of tangent in the term of parameter t then take a point Q as (4t12+34{t_1}^2 + 3,8t13−18{t_1}^3 - 1) now put this point on the equation of tangent on solving find out the relation between t1{t_1} and tt then put the value in point Q.

Complete step-by-step answer:
It is given that the x = 4t2+34{t^2} + 3 , y = 8t3−18{t^3} - 1 , t∈\in R
Differentiate x and y with respect to t .
dxdt=8t\dfrac{{dx}}{{dt}} = 8t and dydt=24t2\dfrac{{dy}}{{dt}} = 24{t^2}
Hence the slope of the curve at point P is
dydx=3t\dfrac{{dy}}{{dx}} = 3t
Equation of tangent = (y−y1)=dydx(x−x1)(y - {y_1}) = \dfrac{{dy}}{{dx}}(x - {x_1})
At point P (4t2+34{t^2} + 3,8t3−18{t^3} - 1)
Equation of tangent become = (y−8t3+1)=t(x−4t2−3)(y - 8{t^3} + 1) = t(x - 4{t^2} - 3)
Hence this tangent meets the curve once again at point Q as (4t12+34{t_1}^2 + 3,8t13−18{t_1}^3 - 1) .
By putting the coordinate in the equation of tangent .
(8t13−1−8t3+1)=3t(4t12+3−4t2−3)(8{t_1}^3 - 1 - 8{t^3} + 1) = 3t(4{t_1}^2 + 3 - 4{t^2} - 3)
8(t13−t3)=24t(t12−t2)8({t_1}^3 - {t^3}) = 24t({t_1}^2 - {t^2})
2(t13−t3)=3t(t12−t2)2({t_1}^3 - {t^3}) = 3t({t_1}^2 - {t^2})
After expanding the equation as a3−b3=(a−b)(a2+b2+ab){a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab) we get
2(t1−t)(t12+t2+tt1)=3t(t1−t)(t1+t)2({t_1} - t)({t_1}^2 + {t^2} + t{t_1}) = 3t({t_1} - t)({t_1} + t)
As t1≠t{t_1} \ne t because it represent same point that is P
2(t12+t2+tt1)=3t(t1+t)2({t_1}^2 + {t^2} + t{t_1}) = 3t({t_1} + t)
2t12+2t2+2tt1=3tt1+t22{t_1}^2 + 2{t^2} + 2t{t_1} = 3t{t_1} + {t^2}
2t12+t2+tt1=02{t_1}^2 + {t^2} + t{t_1} = 0
Separate the t1{t_1}
t12+t2+t12+tt1=0{t_1}^2 + {t^2} + {t_1}^2 + t{t_1} = 0
(t1−t)(t1+t)+t1(t1−t)=0({t_1} - t)({t_1} + t) + {t_1}({t_1} - t) = 0
2t1+t=02{t_1} + t = 0
t1=−t2{t_1} = - \dfrac{t}{2}
After putting the value of t1=−t2{t_1} = - \dfrac{t}{2} in the point Q as (4t12+34{t_1}^2 + 3,8t13−18{t_1}^3 - 1)
After assign the value we get Q (t2+3,−t2−1)({t^2} + 3, - {t^2} - 1)

So, the correct answer is “Option A”.

Note: Always remember that the slope of tangent is dydx\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 dydx\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 t3{t_3} and find relation in between .