Question

The equation of the straight line which is tangent at one point and normal at another point to the curve $y=8t^3-1,x=4t^2+3$ is

• A. $\sqrt{2}x-y=\frac{89\sqrt{2}}{27}-1$
• B. $\sqrt{2}x-y=\frac{89\sqrt{2}}{27}+1$
  
• C. $\sqrt{2}x+y=\frac{89\sqrt{2}}{27}-1$
  
• D. $\sqrt{2}x+y=\frac{89\sqrt{2}}{27}+1$
  

Answer 1

Answer: (B), (C)

$\sqrt{2}x+y=\frac{89\sqrt{2}}{27}-1$
$\sqrt{2}x-y=\frac{89\sqrt{2}}{27}+1$

We have, $y=8t^3-1$ and $x=4t^2+3$

$\Rightarrow \frac{dy}{dt}=24t^2$ and $\frac{dx}{dt}=8t$

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

$\therefore$ tangent at ${}^{\prime }{t}^{\prime }$ is

$y-(8t^3-1)=3t(x-4t^2-3)$ ...(1)

Let the tangent meet the curve again at the point $t^{\prime }$ (say)

$\therefore (8t^{\prime }-1)-(8t^3-1)=3t(4{t^{\prime }}^2+3-4t^2-3)$

$\Rightarrow 8({t^{\prime }}^3-t^3)=3t.4(t^{\prime }-t)(t^{\prime }+t)$

$\Rightarrow 2({t^{\prime }}^2+{t^{\prime }}^2+tr)=3t(t^{\prime }+t)$

$\Rightarrow 2{t^{\prime }}^2-t{t}^{\prime }-t^2=0$

$\Rightarrow (t^{\prime }-t)(2t^{\prime }+t)=0\Rightarrow t^{\prime }=t$ or $\frac{-t}{2}$

$\therefore$ the line (1), which is tangent at the point ${}^{\prime }{t}^{\prime }$ is normal at the point $\frac{-t}{2}$

$\therefore 3t=\frac{2}{3t}\Rightarrow t=\pm \frac{\sqrt{2}}{3}$ Hence, the required lines are

$\sqrt{2}x-y=\frac{89\sqrt{2}}{27}+1$ and $\sqrt{2}x+y=\frac{89\sqrt{2}}{27}-1$