Question

In the curve $x=t^2+3t-8,y=2t^2-2t-5$, at point $(2,-1)$

• A. Length of sub tangent is $\frac{7}{6}$
• B. Slope of tangent $=\frac{6}{7}$
• C. Length of tangent $=\frac{\sqrt{\left(85\right)}}{6}$
• D. Slope of tangent $=\frac{22}{7}$

Answer 1

The correct option is A Length of sub tangent is $\frac{7}{6}$

$x=t^2+3t-8,y=2t^2-2t-5\frac{dx}{dt}=2t+3,\frac{dy}{dt}=4t-2\Rightarrow \frac{dy}{dx}=\left(\frac{dy}{dt}\right)\cdot \left(\frac{dt}{dx}\right)=\left(\frac{4t-2}{2t+3}\right)$

The slope of tangent , $\frac{dy}{dx}$ at $(2,-1)={\left(\frac{dy}{dt}\right)}_{(-2,1)}$

When $x=2,t^2+3t-8=2\Rightarrow t^2+3t-10=0$

$\Rightarrow t^2+5t-2t-10=0\Rightarrow t(t+5)-2(t+5)=0\Rightarrow (t-2)(t+5)=0$

$\Rightarrow t=2,or-5$

$\therefore \frac{dy}{dx}=\left(\frac{4\left(2\right)-2}{2\left(2\right)+3}\right)=\frac{6}{7}$

When $x=2,y=-1$

$\therefore -1=2t^2-2t-5\Rightarrow 2t^2-2t-4=0\Rightarrow 2t^2-4t+2t-4=0$

$\Rightarrow 2t(t-2)+2(t-2)=0\Rightarrow t=2,-1$

So the common value of $t=2$

$\therefore$ Slope of tangent at $(2,-1)=\frac{6}{7}$

Length of tangent $\sqrt{{\left(\frac{y_1}{m}\right)}^2+{\left(y_1\right)}^2}$

Here, $y_1=-1\&m=\frac{6}{7}$

$\therefore$ Length $=\sqrt{{\left(\frac{7}{6}\right)}^2+{(-1)}^2}=\sqrt{\frac{49+36}{6}}=\frac{\sqrt{85}}{6}$

Length of subtangent $\left|\frac{y_1}{\left(\frac{dy}{dt}\right)}\right|=\left|\frac{-1}{\left(\frac{6}{7}\right)}\right|=\frac{7}{6}$