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Show that every homogeneous equation of degree two in x and y, i.e., ax^2 + 2hxy + by^2 = 0, represents a pair of lines passing through the origin, if h^2 – ab ≥ 0. - Mathematics and Statistics

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प्रश्न

Show that every homogeneous equation of degree two in x and y, i.e., ax2 + 2hxy + by2 = 0, represents a pair of lines passing through the origin, if h2 ab 0.

योग
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उत्तर १

Consider a homogeneous equation of the second degree in x and y,

ax2 + 2hxy + by2 = 0   ...(1)

Case I: If b = 0 (i.e., a ≠ 0, h ≠ 0), then the equation (1) reduce to ax2+ 2hxy = 0

i.e., x(ax + 2hy) = 0

Case II: If a = 0 and b = 0 (i.e. h ≠ 0), then the equation (1) reduces to 2hxy = 0, i.e., xy = 0 which represents the coordinate axes and they pass through the origin.

Case III: If b ≠ 0, then the equation (1), on dividing it by b, becomes `a/b x^2 + (2hxy)/b + y^2 = 0`

`therefore y^2 + (2h)/bxy = -a/b x^2`

On completing the square and adjusting, we get `y^2 + (2h)/b xy + (h^2x^2)/b^2 = (h^2x^2)/b^2 - a/b x^2`

`(y + h/bx)^2 = ((h^2 - ab)/b^2)x^2`

`therefore y + h/bx = +- sqrt(h^2 - ab)/b x`

`therefore y = -h/bx +- sqrt(h^2 - ab)/b x`

`therefore y = ((-h +- sqrt(h^2 - ab))/b) x`

∴ Equation represents the two lines `y = ((-h + sqrt(h^2 - ab))/b) x and y = ((-h - sqrt(h^2 - ab))/b) x`

The above equation are in the form of y = mx.

These lines passing through the origin.

Thus, the homogeneous equation (1) represents a pair of lines through the origin, if h2 – ab ≥ 0.

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उत्तर २

Consider a homogeneous equation of degree two in x and y

ax2 + 2hxy + by2 = 0   ...(i)

In this equation at least one of the coefficients a, b or h is non zero. 

We consider two cases.

Case I: If b = 0 then the equqtion

ax2 + 2hxy = 0

x(ax + 2hy) = 0

This is the joint equation of lines x = 0 and (ax + 2hy) = 0.

These lines pass through the origin.

Case II: If b ≠ 0

Multiplying both the sides of equation (i) by b, we get

abx2 + 2hbxy + b2y2 = 0

2hbxy + b2y2 = – abx2

To make LHS a complete square, we add h2x2 on both the sides.

b2y2 + 2hbxy + h2x2 = –abx2 + h2x2

(by + hx)2 = (h2 – ab)x2

`(by + hx)^2 = [(sqrt(h^2 - ab))x]^2`

`(by + hx)^2 - [(sqrt(h^2 - ab))x]^2 = 0`

`[(by + hx) + [(sqrt(h^2 - ab))x]][(by + hx) - [(sqrt(h^2 - ab))x]] = 0`

It is the joint equation of two lines

`(by + hx) + [(sqrt(h^2 - ab))x = 0` and `(by + hx) - [(sqrt(h^2 - ab))x = 0`

`(h + sqrt(h^2 - ab))x + by = 0` and `(h - sqrt(h^2 - ab))x + by = 0`

These lines pass through the origin when h2 – ab > 0.

From the above two cases, we conclude that the equation ax2 + 2hxy + by2 = 0 represents a pair of lines passing through the origin.

shaalaa.com
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2015-2016 (March)
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