**(a**

_{1}x + a_{2}y) + (b_{1}x^{2}+ b_{2}xy + b_{3}y^{2}) + …. + (l_{1}x^{n}+ … + l_{n}y^{n}) = 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

**OP**is

**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 = lim**

_{x0}.

**(2)**

Dividing (1) by x, we obtain

Taking the limit as

**x0**and using

**(2)**, we obtain

**a**

_{1}+ a_{2}m = 0i.e.

**a**

_{1}+ a_{2}Y/X = 0 (∵ Y = mX)i.e.

**a**.

_{1}X + a_{2}Y = 0Thus the equation of the tangent at the origin may be taken as

**a**.

_{1}x + a_{2}y = 0This equation is same as the lowest degree terms in

**(1)**when equated to zero.

If

**a**, then

_{1}= a_{2}= 0**(1)**becomes

**(b**

_{1}x + b_{2}xy + b_{3}y^{2}) + (c_{1}x^{3}+ c_{2}x^{2}y + c_{3}xy^{2}+ c_{4}y^{3}) + …. = 0 (3)Dividing the

**x**

^{2}and taking the limit as

**x0**, we obtain

**b**

_{1}+ b_{2}m + b_{3}m^{2}= 0Or,

**b**

_{1}+ = 0**(∵ Y = mX)**

Or,

**b**.

_{1}X^{2}+ b_{2}XY + b_{3}Y^{2}= 0We may write it as

**b**

_{1}x^{2}+ b_{2}xy + b_{3}y^{2}= 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

**a**and

_{1}= a_{2}= 0**b**, then

_{1}= b_{2}= b_{3}= 0**c**is the equation of the tangent at the origin. Hence we have the following:

_{1}x^{3}+ c_{2}x^{2}y + c_{3}xy^{2}+ c_{4}y^{3}= 0**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

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