# EQUATION OF TANGENT AND NORMAL LINE TO A CONIC SECTION

The equation of tangent at (x1, y1) is obtained from the equation of the curve by replacing

x2 by xx1

y2 by yy1

xy by (xy1 + yx1)/2

x by (x + x1)/2 and

y by (y + y1)/2

Example 1 :

Find the equation of the tangent and normal to the parabola

x2 + x - 2y + 2 = 0

at (1, 2).

Solution :

Equation of tangent at (x1, y1) :

xx1 + (x+x1)/2 − 2(y+y1)/2 + 2 = 0

xx1 + (x+x1)/2 − (y+y1) + 2 = 0

Equation of tangent at (1, 2) :

x(1) + (x+1)/2 − (y+2) + 2 = 0

x + (x+1)/2 − (y+2) + 2 = 0

2x+x+1-2y-4+4  =  0

3x-2y+1  =  0

Equation of normal at (1, 2) :

2x+3y+k  =  0

2(1) + 3(2) + k  =  0

2+6+k  =  0

k  =  -8

2x+3y-8  =  0

Example 2 :

Find the equation of tangent and normal to the ellipse

2x2 + 3y2 = 6

at (3, 0)

Solution :

Equation of tangent at (x1, y1) :

2xx1 + 3yy1 = 6

Equation of tangent at (3, 0) :

2x(3)+3y(0)  =  6

23x  =  6

3x-3  =  0

Equation of normal :

0x-3y+k  =  0

at the point (3, 0) :

0(3)-3(0)+k  =  0

k  =  0

3y =  0

y  =  0

Example 3 :

Find the equation of tangent and normal to the hyperbola

9x2 − 5y2 = 31

at (2, − 1)

Solution :

Equation of tangent at (x1, y1) :

9xx1 - 5yy1 = 31

Equation of tangent at (2, -1) :

9x(2)-5y(-1)  =  31

18x+5y  =  31

Equation of normal :

5x-18y+k  =  0

at the point (2, -1) :

5(2)-18(-1)+k  =  0

10+18+k  =  0

k =  -28

5x-18y-28  =  0

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