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+1 vote
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in Derivatives by (50.2k points)
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The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is:

A. 22/7
B. 6/7
C. -6/7
D. –6

1 Answer

+2 votes
by (48.8k points)
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Best answer

Given equation of curve is x = t2 + 3t – 8, y = 2t2 – 2t – 5

x = t2 + 3t – 8

Differentiating on both sides with respect to t, we get

y = 2t2 – 2t – 5

Differentiating on both sides with respect to t, we get

We know derivative of a constant is 0, so above equation becomes

As given the curve passes through the point (2,-1), substituting these values in given curve equation we get

x = t2 + 3t – 8

⇒ 2= t2 + 3t – 8

⇒ t2 + 3t – 8-2=0

⇒ t2 + 3t – 10=0

Splitting the middle term we get

⇒ t2 + 5t-2t – 10=0

⇒ t(t+ 5) -2(t+5)=0

⇒ (t+ 5) (t-2)=0

⇒ t+5=0 or t-2=0

⇒ t=-5or t=2……….(iii)

y = 2t2 – 2t – 5

⇒ -1=2t2 – 2t – 5

⇒ 2t2 – 2t – 5+1=0

⇒ 2t2 – 2t – 4=0

Taking 2 common we get

⇒ t2 – t – 2=0

Splitting the middle term we get

⇒ t2 – 2t +t– 2=0

⇒ t(t– 2)+1(t– 2)=0

⇒ (t– 2)(t+1)=0

⇒ (t– 2)=0 or (t+1)=0

⇒ t=2 or t=-1……..(iv)

So from equation (iii) and (iv), we can see that 2 is common

So t=2

So the slope of the tangent at t=2 is given by

Therefore, the slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is 6/7.

So the correct option is option B.

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