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Question
Show that the homogeneous equation of degree 2 in x and y represents a pair of lines passing through the origin if h2 − ab ≥ 0.
Solution
Consider the homogeneous equation of degree two in x and y
ax2 +2hxy + by2 = 0 .......(i)
Consider two cases b = 0 and b ≠ 0.
These two cases are exhaustive.
Case I:
If b = 0 then the equation ax2 + 2hxy = 0
∴ x(ax + 2hy) = 0, which is the combined equation of lines x = 0 and ax + 2hy = 0.
We observe that these lines pass through the origin.
Case II:
If b ≠ 0, then we multiply equation (i) by b
abx2 + 2hbxy + b2y2 = 0
∴ b2y2 + 2hbxy = – abx2
To make L.H.S. complete square we add h2x2 to both sides.
b2y2 + 2hbxy + h2x2 = h2x2 – abx2
∴ (by + hx2 = (h2 – ab)x2
∴ `("b"y + "h"x)^2 = (sqrt("h"^2 - "ab"))^2 x^2, "as" "h"^2 - "ab" ≥ 0`
∴ `("b"y + "h"x)^2 - (sqrt("h"^2 - "ab"))^2 x^2` = 0
∴ `("b"y + "h"x + sqrt("h"^2 - "ab"x))("b"y + "h"x - sqrt("h"^2 - "ab" x))` = 0
∴ `[("h" + sqrt("h"^2 - "ab"))x + "b"y]*[("h" - sqrt("h"^2 - "ab"))x + "b"y]`= 0,
which is the combined equation of lines
`("h" + sqrt("h"^2 - "ab"))x + "b"y` = 0 and ("h" - sqrt("h"^2 - "ab"))x + "b"y` = 0
As b ≠ 0, we can write these equations in the form
= m1x and y = m2x,
Where m1 = `(-"h" - sqrt("h"^2 - "ab"))/"b"` and m2 = `(-"h" + sqrt("h"^2 - "ab"))/"b"`
We observe that these lines pass through the origin.
∴ From the above two cases, we conclude that the equation ax2 + 2hxy + by2 = 0 represents a pair of lines passing through the origin, if h2 − ab ≥ 0.
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