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

\frac{22}{7}

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\frac{6}{7}

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\frac{7}{6}

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- \frac{6}{7}

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Solution

The correct option is B $\frac{6}{7}$
$(x, y) = (2, - 1)$
$t^{2} + 3 t - 8 = 2 \Rightarrow t^{2} + 3 t - 10 = 0$
$(t + 5) (t - 2) = 0 \Rightarrow t = - 5, 2$

$2 t^{2} - 2 t - 5 = - 1 \Rightarrow 2 t^{2} - 2 t - 4 = 0$
$t^{2} - t - 2 = 0 \Rightarrow (t - 2) (t + 1) = 0$
$\Rightarrow t = 2, - 1$
Therefore, $t = 2$
$\frac{d y}{d x} = \frac{4 t - 2}{2 t + 3} = \frac{6}{7}$

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Similar questions

The slope of the tangent to the curveat the point (2, −1) is

(A) (B) (C) (D)

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

(a) 22/7
(b) 6/7
(c) −6
(d) none of these

Choose the correct answer in the following questions :
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
(a) $\frac{22}{7}$ (b) $\frac{6}{7}$
(c) $\frac{7}{6}$ (d) $\frac{- 6}{7}$

Q. The slope of tangent to the curve

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

at the point

(2, - 1)

is :

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) $\frac{7}{6}$
(d) $- \frac{6}{7}$

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

The correct option is B 67(x,y)=(2,−1) t2+3t−8=2⇒t2+3t−10=0 (t+5)(t−2)=0⇒t=−5,2 2t2−2t−5=−1⇒2t2−2t−4=0 t2−t−2=0⇒(t−2)(t+1)=0 ⇒t=2,−1 Therefore, t=2 dydx=4t−22t+3=67

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