Let the equation of the curve passing through the origin be
(a1x + a2y) + (b1x2 + b2xy + b3y2) + …. + (l1xn + … + lnyn) = 0 (1)
Let P (x, y) be any point on the curve. The slope of the chord OP is . Thus the equation of the chord OPis
Y = .
As x0, the chord OP becomes the tangent at O and so the equation of the tangent at O is
Y = mX, where m = limx0 . (2)
Dividing (1) by x, we obtain
Taking the limit as x0 and using (2), we obtain
a1 + a2m = 0
i.e. a1 + a2 Y/X = 0 (∵ Y = mX)
i.e. a1X + a2Y = 0.
Thus the equation of the tangent at the origin may be taken as
a1x + a2y = 0.
This equation is same as the lowest degree terms in (1) when equated to zero.
If a1 = a2 = 0, then (1) becomes
(b1x + b2xy + b3y2) + (c1x3 + c2x2y + c3xy2 + c4y3) + …. = 0 (3)
Dividing the x2 and taking the limit as x0, we obtain
b1 + b2m + b3m2 = 0
Or, b1 + = 0 (∵ Y = mX)
Or, b1X2 + b2XY + b3Y2 = 0.
We may write it as
b1 x2 + b2xy + b3y2 = 0. (4)
which represents a pair of tangents at this origin.
The equation (4) is same as the lowest degree terms in (3) when equated to zero.
Similarly, it can be shown that if a1 = a2 = 0 and b1 = b2 = b3 = 0, then c1x3 + c2x2y + c3xy2 + c4y3 = 0is the equation of the tangent at the origin. Hence we have the following:
Rule: The tangents at the origin are given by equating to zero the lowest degree terms in the equation of the given curve according to the math homework help experts