The slope of the tangent to the curve x=t2+3t−8, y=2t2−2t−5 at the point (2, -1), is

A
$\frac{22}{7}$
B
$\frac{6}{7}$
C
-6
D
none of these

Solution:

For the point $(2, - 1)$ on the curve

$x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5,$ we have

$\begin{aligned} t^{2} + 3 t - 8 = 2 and 2 t^{2} - 2 t - 5 = - 1 \\ \Rightarrow t^{2} + 3 t - 10 = 0 and 2 t^{2} - 2 t - 4 = 0 \\ \Rightarrow (t + 5) (t - 2) = 0 and (t - 2) (t + 1) = 0 \\ \Rightarrow t = 2 \end{aligned}$

Now, $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{d x} = \frac{4 t - 2}{2 t + 3} \Rightarrow {(\frac{d y}{d x})}_{t = 2} = \frac{4 \times 2 - 2}{2 \times 2 + 3} = \frac{6}{7}$

The slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at the point $(2, - 1)$ , is

Solution:

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