Question

The slope of the tangent at $(1,2)$ to the curve $x=t^2-7t+7$ and $y=t^2-4t-10$, is

• A. $\frac{8}{5}$
• B. $\frac{5}{8}$
• C. $-\frac{8}{5}$
• D. $-\frac{5}{8}$

Answer 1

Answer: (A)

We have,
$x=t^2-7t+7,y=t^2-4t-10$
Put $x=1$
$1=t^2-7t+7$
$\Rightarrow t^2-7t+6=0$
$\Rightarrow (t-6)(t-1)=0$
$\Rightarrow t=1,6$
Put $y=2$
$2=t^2-4t-10$
$\Rightarrow t^2-4t-12=0$
$(t-6)(t+2)=0$
$\Rightarrow t=-2,6$
Hence, $t=6$ satisfies both equations.
Now, $\frac{dx}{dt}=2t-7$
$\Rightarrow {\left(\frac{dx}{dt}\right)}_{t=6}=5$
and $\frac{dy}{dt}=2t-4$
$\Rightarrow {\left(\frac{dy}{dt}\right)}_{t=6}=8$
$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{8}{5}$