HSC Science (General) 12th Board ExamMaharashtra State Board
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# 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. - HSC Science (General) 12th Board Exam - Mathematics and Statistics

ConceptPair 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 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.

#### Solution 2

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.

Is there an error in this question or solution?

#### APPEARS IN

2015-2016 (March) (with solutions)
Question 3.1.3 | 3 marks
2013-2014 (March) (with solutions)
Question 3.1.1 | 3 marks
2017-2018 (March) (with solutions)
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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: null - Pair of Lines Passing Through Origin - Homogenous Equation. For the courses HSC Science (General) , HSC Science (Computer Science), HSC Science (Electronics), HSC Arts
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