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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) $\frac{-6}{7}$
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) $\frac{-6}{7}$
Solution:
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Verified Answer
(b) The curve is $x=t^2+3 t-8, y=2 t^2-2 t-5 \ldots$ (i)
Put $\mathrm{x}=2, \Rightarrow(\mathrm{t}+5)(\mathrm{t}-2)=0$
put $t=2$ in $y=2 t^2-2 t-5=8-4-5=-1$
At $x=2, y=-1, t=2$
Differentiating (i) $\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t}+3, \frac{\mathrm{dy}}{\mathrm{dt}}=4 \mathrm{t}-2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{4 \mathrm{t}-2}{2 \mathrm{t}+3} ; \quad \text { At } \mathrm{t}=2, \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6}{7}$
Put $\mathrm{x}=2, \Rightarrow(\mathrm{t}+5)(\mathrm{t}-2)=0$
put $t=2$ in $y=2 t^2-2 t-5=8-4-5=-1$
At $x=2, y=-1, t=2$
Differentiating (i) $\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t}+3, \frac{\mathrm{dy}}{\mathrm{dt}}=4 \mathrm{t}-2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{4 \mathrm{t}-2}{2 \mathrm{t}+3} ; \quad \text { At } \mathrm{t}=2, \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6}{7}$
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