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
2√3x = 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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