Q222 / 270 · NDA 2012 (Phase 1) · Mathematics

What is the slope of the tangent to the curve $x=t^2+$ $3 t-8, y=2 t^2-2 t-5$ at $t=2 ?$

  • A
    7/6
  • Correct answer
    B
    6/7
  • C
    1
  • D
    5/6

Solution

$\begin{aligned} & \frac{d x}{d t}=2 t+3, \frac{d y}{d t}=4 t-2 \\ & \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{4 t-2}{2 t+3} \\ & \text {slope at } t=2=\frac{d y}{d x} / t=2 \\ & =\frac{4(2)-2}{2(2)+3}=\frac{6}{7} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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