First slide

Tangents and normals

Question

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

Moderate

A

7/6
B

2/3
C

3/2
D

6/7

Solution

$We first determine the value of t corresponding to the given$ $values of x and y. From t^{2} + 3 t - 8 = 2, we get$
$t = 2, - 5, and from 2 t^{2} - 2 t - 5 = - 1, we get t = 2, - 1 .$
Hence to the given point there corresponds thevalue t = 2.
Therefore, the slope of the tangent at $(2, - 1)$ is
$y' |_{t = 2} = {\frac{d y / d t}{d x / d t}|}_{t = 2} = {\frac{4 t - 2}{2 t + 3}|}_{t = 2} = \frac{6}{7}$

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