The slope of the tangent to the curve x=t2+3t−8y=2t2−2t

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

A. $\frac{7}{6}$

$\frac{5}{6}$

$\frac{6}{7}$

$1$

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

We have,

$\frac{d x}{d t} = 2 t + 3 a n d \frac{d y}{d t} = 4 t - 2 \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{4 t - 2}{2 t + 3}$
Thus, slope of the tangent to the curve at the point t = 2 is
${[\frac{d y}{d x}]}_{t - 2} = \frac{4 (2) - 2}{2 (2) + 3} = \frac{6}{7}$
Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer

The slope of the tangent to the curve x=t2+3t−8y=2t2−2t

Right Answer is: C

SOLUTION

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Related Questions
If m be the slope of a tangent to the curve e2y=1+4x2 t
For the parabola y2=4 a x the ratio of the sub-tangent
The sub-normal at any point of the curve x2y2=a2x2−a2 v
The two curves x3−3xy2+2=0 and 3x2y−y3−2=0 Cut at righ