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  • If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3 − 1, t <img alt="\varepsilon" src=

If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3 − 1, t \varepsilon R, meets the curve again at a point Q, then the coordinates of Q are :

Option: 1

( t2 + 3,  −t3 −1 )


Option: 2

( 4t2 + 3,  −8t3 −1 )


Option: 3

( t2 + 3,  t3 − 1 )


Option: 4

( 16t2 + 3, −64t3 −1 )


Answers (1)

best_answer

As learnt in concept @MNOP

x= 4t^2 +3\:and\: y=8t^3-1

P(4t^2+3, 8t^3-1) and Q(4t_1^2+3, 8t_1^3-1)

at P; \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{24t^2}{8t} =3t

Tangent at P is

y-(8t^3-1)= 3t[x-(4t^2+3)]

Put \: (4t^2_1+3, 8t^2_1-1)

We get

8t^3_1- 8t^3 = 3t(4t^2_1- 4t^2)

2(t^2_1+t_1t+t^2)= 3(t_1+t)

(t_1-t)(2t_1+t)=0

Hence t_1 =\frac{t}{2}

Hence, Q = ( t2 + 3,  t3 −1 )

Posted by

Deependra Verma

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