Question

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

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

Answer 1

Answer: (A)

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