Question

If a straight line is tangent to one point and normal to another point on the curve $y=8t^3-1,x=4t^2+3$, then the equation of such line is

• A. $x-y=\sqrt{2}$
• B. $\sqrt{2}x-y=\left[\frac{\left(89\sqrt{2}\right)}{27}\right]+1$
• C. $x-y=89\sqrt{2}$
• D. None of these

Answer 1

The correct option is B

$\sqrt{2}x-y=\left[\frac{\left(89\sqrt{2}\right)}{27}\right]+1$

Explanation for the correct option:

Step 1. Finding the equation of the line:

Given Point of curve $y=8t^3-1,x=4t^2+3$

Differentiate both with respect to $t$

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

Let the two points be $P\left(4{t_1}^2+3,8{t_1}^3-1\right)$ and $Q\left(4{t_2}^2+3,8{t_2}^3-1\right)$

Now $\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}$

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

$\Rightarrow$ $\frac{dy}{dx}=3t_1=\mathrm{slope}\mathrm{of}\mathrm{tangent}at{t}_1$

Step 2. equation of tangent at P is

$y-\left(8{t_1}^3-1\right)=3t_1\left[x-\left(4{t_1}^2+3\right)\right]$

$\Rightarrow$ $8\left({t_2}^3-{t_1}^3\right)=3t_1·4\left({t_2}^2-{t_1}^2\right)$ [As normal pass via point $t_2$]

$\Rightarrow$ $2\left({t_1}^2-{t_2}^2+t_2{t}_1\right)=3t_1\left(t_2+t_1\right)$ $\left[\because t_2\ne t_1\right]$

$\Rightarrow$ $2{t_2}^2-t_2{t}_1-{t_1}^2=0$

$\Rightarrow$ $\left(t_2-t_1\right)\left(2t_2+t_1\right)=0$

$$\begin{gathered} \because t_2\ne t_1 \\ \therefore t_1=-2t_2—\left(1\right) \end{gathered}$$

$\therefore$ Slope of normal at $Q=\frac{1}{3t_2}$

Since tangent at P is normal Q

$\therefore$ $3t_1=\frac{1}{3t_2}$

$\Rightarrow$$9t_2{t}_1=-1—\left(2\right)$

Step 3. From $\left(1\right)$ and $\left(2\right)$, we get

$9t_1\left(-\frac{t_1}{2}\right)=-1$

$\Rightarrow$ ${t_1}^2=\frac{2}{9}$

$\Rightarrow$ $t_1=\frac{\sqrt{2}}{3}$

$\therefore$ required line is $y-\left(8\times \frac{2\sqrt{2}}{27}-1\right)$$=3\frac{\sqrt{2}}{3}\left[x-\left(4\times \frac{2}{9}+3\right)\right]$

$\Rightarrow$ $y-\frac{16}{27\sqrt{2}}+1=\sqrt{2}x-\frac{35}{9\sqrt{2}}$

$\Rightarrow$ $\begin{array}{rcl}\sqrt{2}x-y& =& \frac{35}{9\sqrt{2}}-\frac{16}{27\sqrt{2}}+1\end{array}$

$\Rightarrow$ $\sqrt{2}x-y\begin{array}{rcl}& =& \end{array}\frac{89\sqrt{2}}{27}+1$

Hence, option (B) is correct .