Find the equation of the tangent and normal to the parabola $y^{2}=12x $ at $ (3 , -6 )$

Answer 1

Toolbox:

• The equation of the tangent at $(x_1,y_1)$ to a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is given by $Axx_1+\large\frac{B}{2}$$(xy_1+x_1y)+Cyy_1+D\large\frac{(x+x_1)}{2}+$$E\large\frac{(y+y_1)}{2}$$+F=0$
• The normal is the perpendicular to this line that passes through $(x_1,y_1)$
• Any line perpendicular to $ax+by+c=0$ is of the form $bx-ay+k=0$.This can be used to find the equation of the normal.

Step 1:

Equation of the tangent and normal to $y^2=12x$ at $(3,-6)$

The equation of the tangent is given by $yy_1=12\large\frac{(x+x_1)}{2}$

Where $(x_1,y_1)=(3,-6)$

(i.e) $y(-6)=6(x+3)$

$\Rightarrow 6x+6y+18=0$

$\Rightarrow x+y+3=0$

Step 2:

The normal is $\perp$ to the tangent.So its equation is of the form $x-y+k=0$.It passes through $(3,-6)$

Therefore $3+6+k=0$

$\Rightarrow k==-9$

The normal is $x-y-9=0$

Step 3:

The equation of tangent : $x+y+3=0$

The equation of normal : $x-y-9=0$