Account
It's free!

User



Login
Register


      Forgot password?
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution - How 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 Origin If h^2-ab≥0. - Pair of Straight Lines - Pair of Lines Passing Through Origin - Homogenous Equation

Question

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 origin if h2ab0.

Solution 1

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

`ax^2+2hxy+by^2=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^2-1/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.

Solution 2

Consider a homogeneous equation of degree two in x and y

`ax^2+2hxy+by^2=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

`ax^2+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

`abx^2+2hbxy+b^2y^2=0 `

`2hbxy+b^2y^2=-abx^2`

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

`b^2y^2+2hbxy+h^2y^2=-abx^2+h^2x^2`

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

`(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 `ax^2+2hxy+by^2=0` represents a pair of lines passing through the origin.

 

 

Is there an error in this question or solution?

APPEARS IN

2015-2016 (March)
Question 3.1.3 | 3 marks
2014-2015 (October)
Question 2.1.2 | 3 marks
2013-2014 (March)
Question 3.1.1 | 3 marks
2017-2018 (March)
Question 2.2.1 | 4 marks
Solution for question: How 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 Origin If h^2-ab≥0. concept: Pair of Straight Lines - Pair of Lines Passing Through Origin - Homogenous Equation. For the courses HSC Arts, HSC Science (Computer Science), HSC Science (Electronics), HSC Science (General)
S