Question

If a line is tangent at one point and normal at another point on the curve $x=4t^2+3,y=8t^3-1$, then slope(s) of such a line is/are

• A. $-2$
• B. $2$
• C. $\sqrt{2}$
• D. $-\sqrt{2}$

Answer 1

Answer: (C), (D)

$-\sqrt{2}$

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

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 $8(t_1^3-t^3)=3t\times 4(t_1^2-t^2)$
$\Rightarrow (t_1-t{)}^2(t+2t_1)=0$
$\Rightarrow t=-2t_1(\text{as}t\ne t_1$)

Also, ${\left(\frac{dy}{dx}\right)}_P=-\frac{1}{{\left(\frac{dy}{dx}\right)}_Q}$
$\Rightarrow 3t=-\frac{1}{3t_1}$
$\Rightarrow t{t}_1=-\frac{1}{9}$
$\Rightarrow -2t_1^2=-\frac{1}{9}$
$\Rightarrow t_1=\pm \frac{1}{3\sqrt{2}}$
Therefore, slope of line is $-\frac{1}{3t_1}=\pm \sqrt{2}$