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Question

If tangent at point p, with parameter t, on the curve x=4t2+3,y=8t31,tϵR, meets the curve again at point Q, then the coordinates of Q are:-


A
t2+3,-t31
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B
4t2+3,8t21
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C
t2+3,t31
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D
16t2+3,64t31
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Solution

The correct option is A t2+3,-t31
Given,

P(x=4t2+3,y=8t31)

dydx=dydtdydt

=24t28t=3t

Tangent at P is given by, (y8t3+1)=3t(x4t23)

Let,

Q(4t21+3,8t311)

Therefore Q satisfies the equation of tangent at P,

8t3118t3+1=3t(4t21+34t23)

8t318t3=3t(4t214t2)

8(t1t)(t2+tt1+t21)=3t×4(t1t)(t1+t)

2(t2+tt1+t21)=3t(t1+t)

2t21tt1+t2=0

(t1t)(2t1+t)=0

t1=t2

Q=(4t21+3,8t311)=(t2+3,t31)

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