Question

Find the equation of the tangent and normal to the curves
(i) $y=x^2-4x-5$ at $x=-2$
(ii) $y=x-\sin x\cos x$
(iii) $y=2{\sin }^{2}3x$ at $x=\frac{\pi }{6}$

Answer 1

$\left(i\right)$

$y=x^2-4x-5$

at $x=-2$

$y=(-2{)}^2+8-5=7$

$\frac{dy}{dx}=2x-4$

at $x=-2,\frac{dy}{dx}=-8$

eq of tangent :

$(y-7)=-8(x+2)$

$\Rightarrow y+8x+9=0$

$\frac{eqofnormal}{(y-7)=\frac{1}{8}(x+2)}$

$\Rightarrow x-8y+58=0$

$\left(ii\right)$

$y=x-\sin xwsx$

$at$x= \dfrac{x}{6}$

$y=\frac{\pi }{6}-\frac{1}{2}.\frac{\sqrt{3}}{2}$

$=\frac{\pi }{6}-\frac{\sqrt{3}}{4}$

$\frac{dy}{dx}=1-{\cos }^{2}x+{\sin }^{2}x$

$=2{\sin }^{2}x$

at $x=\frac{\pi }{6},\frac{dy}{dx}=2.\frac{1}{4}=\frac{1}{2}$

eq of tangent :

$(y-\frac{\pi }{6}+\frac{\sqrt{3}}{4})=\frac{1}{2}(x-\frac{\pi }{6})$

eq of normal

$(y-\frac{\pi }{6}+\frac{\sqrt{3}}{4})=-2(x-\frac{\pi }{6})$

$\left(iii\right)$

$y=2{\sin }^{2}3x$

at $x=\frac{\pi }{6}$

$y=2{\sin }^2\frac{\pi }{2}=2$

$\frac{dy}{dx}=4\sin 3x,\cos 3x$

at $x=\frac{\pi }{6},\frac{dy}{dx}=0$.

eq of tangent :

$(y-2)=0$

$\Rightarrow y=2$

eq of normal

$x-\frac{\pi }{6}=0$

$\Rightarrow x=\frac{\pi }{6}$