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Find the equation of the tangent and normal to the parabola $x^{2}=9y$ at $(3 , 1 )$

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  • 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 $x^2=9y$ at $(-3,1)$
The equation of the tangent is given by $xx_1=9\large\frac{(y+y_1)}{2}$
Where $(x_1,y_1)=(-3,1)$
Therefore the tangent is $-3x_1=\large\frac{9(y+1)}{2}$
(i.e) $6x+9y+9=0$
The above equation is divided by 3 we get
Step 2:
The normal is $\perp$ to the tangent .So its equation is of the form $3x-2y+k=0$
It passes through $(-3,1)$
$\Rightarrow k=11$
The normal is $3x-2y+11=0$
Step 3:
The equation of tangent : $2x+3y+3=0$
The equation of normal : $3x-2y+11=0$
answered Jun 20, 2013 by sreemathi.v

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