Question

Find the slope of the line which is tangent at one point and normal at another point on the curve $x=4t^2+3,y=8t^3-1$.

Answer 1

$x=4t^2+3,y=8t^3-1$
Differentiate w.r.t. $x$
$\frac{dx}{dt}=8t,\frac{dy}{dt}=24t^2$

$\therefore \frac{dy}{dx}=\frac{24t^2}{8t}=3t$

Let the tangent at $P(4t^2+3,8t^3-1)$ be normal at $Q(4t_1^2+3,8t_1^3-1)$

Equation of tangent at $P$ is $y-(8t^3-1)=3t(x-4t^2-3)$
This tangent passes through $Q$, then
$(8t_1^3-1)-(8t^3-1)=3t\left[\right(4t_1^2+3)-(4t^2+3\left)\right]$
$\Rightarrow 8(t_1^3-t^3)=3t\times 4(t_1^2-t^2)$
$\Rightarrow 2(t_1^2+t_{1}t+t^2)=3t(t_1+t)$
$\Rightarrow 2t_1^2-t{t}_1-t^2=0$
$\Rightarrow (t_1-t)(2t_1+t)=0$
$\Rightarrow t=-2t_1(\because t\ne t_1)$
$\Rightarrow t_1=-\frac{t}{2}$

Now, ${\left(\frac{dy}{dx}\right)}_t=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{t_1}}$
$\Rightarrow 3t=-\frac{1}{3t_1}$
$\Rightarrow t{t}_1=-\frac{1}{9}$
$\Rightarrow t\times (-\frac{t}{2})=-\frac{1}{9}$
$\Rightarrow t^2=\frac{2}{9}$
$\Rightarrow t=\pm \frac{\sqrt{2}}{3}$


$\therefore$ Slope of the line : $3t=\pm \sqrt{2}$