## Friday, October 14, 2011

### Tangents at the Origin math homework help

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