Comment
Share
Q)

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

Comment
A)
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 : $2x+y-1=0$