What is the slope of the line normal to the tangent line of f(x) = 2x^2-xsqrt(x^2-x)  at  x= 2 ?

Feb 9, 2017

$\text{The Reqd. Slope=} \frac{16 + 5 \sqrt{2}}{103}$.

Explanation:

Recall that, $f ' \left(2\right)$ gives the slope of the Tangent Line to the

Curve C : $y = f \left(x\right) = 2 {x}^{2} - x \sqrt{{x}^{2} - x}$

at the pt. $P \left(2 , f \left(2\right)\right) = P \left(2 , 8 - 2 \sqrt{2}\right)$.

As Normal at pt. $P$ is $\bot$ to the tgt., its slope is given by,

$\frac{1}{f ' \left(2\right)} , \mathmr{if} f ' \left(2\right) \ne 0$.

Now, $f \left(x\right) = 2 {x}^{2} - x \sqrt{{x}^{2} - x} = 2 {x}^{2} - \sqrt{{x}^{4} - {x}^{3}} .$

$\therefore f ' \left(x\right) = 2 \left(2 x\right) - \frac{1}{2 \sqrt{{x}^{4} - {x}^{3}}} \cdot \left({x}^{4} - {x}^{3}\right) '$

$= 4 x - \frac{1}{2 \sqrt{{x}^{4} - {x}^{3}}} \cdot \left(4 {x}^{3} - 3 {x}^{2}\right)$

$= 4 x - \frac{{x}^{2} \left(4 x - 3\right)}{2 x \sqrt{{x}^{2} - x}}$

$\therefore f ' \left(x\right) = 4 x - \frac{x \left(4 x - 3\right)}{2 \sqrt{{x}^{2} - x}}$

$\Rightarrow f ' \left(2\right) = 8 - \frac{2 \left(8 - 3\right)}{2 \sqrt{4 - 2}} = 8 - \frac{5}{\sqrt{2}} = \frac{8 \sqrt{2} - 5}{\sqrt{2}} \ne 0.$

Hence, the Reqd. Slope $= \frac{\sqrt{2}}{8 \sqrt{2} - 5} = \frac{\sqrt{2} \left(8 \sqrt{2} + 5\right)}{{\left(8 \sqrt{2}\right)}^{2} - {5}^{2}}$

$= \frac{16 + 5 \sqrt{2}}{103}$.