Question

The slope of the tangent to the curve x = t² + 3t – 8,

y = 2t² – 2t – 5 at the point (2, –1) is –

• A. 6/7
• B. –6
• C. 22/7
• D. None of these

Answer 1

Answer: (A)

6/7

Answer 2

By (2, – 1) x = 2 = t² + 3t – 8 Ž t² + 3t – 10 = 0

Ž (t + 5) (t – 2) = 0 Ž t = 2, – 5

y = – 1 = 2t² – 2t – 5 Ž 2t² – 2t – 4 = 0

Ž (t – 2) (t + 1) = 0 Ž t = 2, – 1

so t = 2 Ž New slope = $\frac{dy}{dx}$= $\frac{4t–2}{2t + 3}$

= $\frac{4 \times 2–2}{2 \times 2 + 3}$= $\frac{6}{7}$