Question

Find the equations of tangent and normal to the curves at the indicated point on it.
(a) $y=x^2+4x+1$ at $\left(-1,-2\right)$
(b) $2x^2+3y^2-5=0$ at $\left(1,1\right)$
(c) $x=a{\cos }^3\theta ,y=a{\sin }^3\theta at\theta =\frac{\pi }{4}$

Answer 1

$\left(a\right)y=x^2+4x+1$

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

${\left[\frac{dy}{dx}\right]}_{(-1,-2)}=2\times -1+4=-2+4=2$

$\therefore$Slope of the tangent$=2$

Equation of the tangent is $y-(-2)=2(x-(-1))$

$y+2=2(x+1)$

$y+2=2x+2$

$\therefore y=2x$ is the equation of the tangent.

Slope of the normal$=-\frac{1}{2}$

Equation of the normal is $y-(-2)=\frac{-1}{2}(x-(-1))$

$2y+4=-(x+1)$

$2y+4+x+1=0$

$\therefore x+2y+5=0$ is the equation of the normal.

$\left(b\right)2x^2+3y^2-5=0$

$\Rightarrow 4x+6y\frac{dy}{dx}=0$

$\Rightarrow 6y\frac{dy}{dx}=-4x$

$\Rightarrow \frac{dy}{dx}=\frac{-4x}{6y}=\frac{-2x}{3y}$

${\left[\frac{dy}{dx}\right]}_{(1,1)}=\frac{-2\times 1}{3\times 1}=-\frac{2}{3}$

$\therefore$Slope of the tangent$=-\frac{2}{3}$

Equation of the tangent is $y-1=-\frac{2}{3}(x-1)$

$3y-3=-2x+2$

or $2x+3y-5=0$ is the equation of the tangent.

The slope of the normal$=\frac{3}{2}$

Equation of the normal is $y-1=\frac{3}{2}(x-1)$

$2y-2=3x-3$

$3x-2y-1=0$

$\therefore 3x-2y-1=0$ is the equation of the normal.

$\left(c\right)x=a{\cos }^3\theta ,y=a{\sin }^3\theta$

$\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta ,\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{\theta =\frac{\pi }{4}}=-\tan \frac{\pi }{4}=-1$

$\therefore$Slope of the tangent$=-1$

Equation of the tangent is $y-a{\sin }^3\theta =-1(x-a{\cos }^3\theta )$

or $x+y=a({\sin }^3\theta +{\cos }^3\theta )$

or $x+y=a(\sin \theta +\cos \theta )({\sin }^2\theta +{\cos }^2\theta -\sin \theta \cos \theta )$

or $x+y=a(\sin \theta +\cos \theta )(1-\sin \theta \cos \theta )$

$\therefore$Slope of the normal$=1$

Equation of the normal is $y-a{\sin }^3\theta =1(x-a{\cos }^3\theta )$

or $x-y=a({\sin }^3\theta -{\cos }^3\theta )$

or $x-y=a(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )$

or $x-y=a(\sin \theta -\cos \theta )(1+\sin \theta \cos \theta )$