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

1 Answer

  • 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 : $x-y-9=0$
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