Question

a.Find the equation of the normal to the curve $y=|x^2-\left|x\right||atx=-2.$
b. Find the equation of tangent to the curve
$y={sin}^{-1}\frac{2x}{1+x^2}atx=\sqrt{3}$

Answer 1

We have,

$y=|x^2-\left|x\right||$

On differentiating and we get,

$\frac{dy}{dx}=2x-1$

At point $(x=-2)$

Then,

${\left(\frac{dy}{dx}\right)}_{(x=-2)}=|2\times (-2)-1|$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x=-2)}=|-5|$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x=-2)}=5$

\end{align}$

Now the equation of normal is

$y-y_1=m(x-x_1)$

$y-3=-\frac{1}{m}(x+2)\therefore fornormal$

$y-3=-\frac{1}{5}(x+2)$

$5y-15=-(x+2)$

$5y-15=-x-2$

$5y-15-x+2=0$

$-x+5y-13=0$

$x-5y+13=0$

Hence, this is the answer.