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Find the equation of the tangent and normal to the parabola $x^{2}+2x-4y+4=0$ at $(0 , 1 )$

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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}$$+F=0$
  • The normal is the perpendicular to this line that passes through $(x_1,y_1)$
  • Any line perpendicular to $ax+by+c=0$ is of the form $bx-ay+k=0$.This can be used to find the equation of the normal.
Step 1:
Tangent and normal at $(0,1)$ to $x^2+2x-4y+4=0$
The tangent is at $(x_1,y_1)$ is
$xx_1+\large\frac{2(x+x_1)}{2}-$$4\large\frac{(y+y_1)}{2}$$+4=0$
Here $(x_1,y_1)=(0,1)$
The tangent is $0+x-2(y+1)+4=0$
(i.e) $x-2y+2=0$
Step 2:
The normal is $\perp$ to the tangent and its equation is of the form $2x+y+k=0$.It passes through $(0,1)$.
Therefore $0+1+k=0$
$\Rightarrow k=-1$
The normal is $2x+y-1=0$
Step 3:
The equation of tangent : $x-2y+2=0$
The equation of normal : $2x+y-1=0$
answered Jun 20, 2013 by sreemathi.v
 

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