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 h²- ab ≥ 0.​

Answer 1

Consider a homogeneous equation of degree two in x and y, ${{ax}^{2} + 2hxy + {hy}^{2} = 0}$ ..(1) ‎

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

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We consider two cases. ‎

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Case l: If b = 0,

then the equation of lines are x = 0 and (ax + 2hy) = 0.

These lines passes through the origin.

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Case ll: $\sf{b \: \neq \: 0}$,

Multiplying both the sides of equation (1) by b, we get

${abx}^{2} + 2hxyb \: + {hby}^{2} = 0$

$\Rightarrow{{hby}^{2} + 2hxyb = {-abx}^{2}}$‎

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To make L.H.S a complete square, we add ${h}^{2}{x}^{2}$ on both the sides.‎

$\Rightarrow{{hby}^{2} + 2hxyb + {h}^{2} {x}^{2} = {-abx}^{2} + {h}^{2} {x}^{2}}$

$\Rightarrow{{by + hx}^{2} = ({h}^{2} - ab){x}^{2}}$

$\Rightarrow{{by + hx}^{2} = {[(\sqrt{{h}^{2} - ab})x]}^{2} }$

$\Rightarrow{{by + hx}^{2} - {[(\sqrt{{h}^{2} - ab})x]}^{2} = 0}$

$\Rightarrow[(by + hx) - {(\sqrt{{h}^{2} - ab})}x] \times [(by + hx) + {(\sqrt{{h}^{2} - ab})}x] = 0$‎

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It is the joint equation of two lines,

$(by + hx) - {(\sqrt{{h}^{2} - ab})}x = 0$ and

$(by + hx) + {(\sqrt{{h}^{2} - ab})}x = 0$‎

‎

i.e, $(h + {(\sqrt{{h}^{2} - ab})}x + by = 0$ and

$(h - {(\sqrt{{h}^{2} - ab})}x + by = 0$‎

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Therefore, These lines passes through the origin.

Answer 2

Step-by-step explanation:

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$\huge{\underline{\mathrm{Question}}}$

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 h²- ab ≥ 0.

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$\huge{\underline{\mathrm{Solution\: :}}}$

Let two lines y=m1,x and y=m2x passess through origin.

$\sf\red{⟼y−m1x=0 and y−m2x=0 }$

$\sf\green{⟼(y−m1x)(y−m2x)=0 }$

$\sf\orange{∴y2−(m1+m2)xy+m1m2x2=0; }$

⠀⠀⠀⠀⠀⠀[ Where m1 and m2 are slopes of lines. ]

$\sf\blue{∴y2−(sum\: of \:slopes)xy+(product\: of \:slopes)x2=0 }$

[It represents pair of straight lines passing through origin.]

Now,

$\sf\pink{ax2+2hxy+by2=0 }$

$\sf\red{⟼y2+b2hxy+bax2=0}$

$\sf\blue{⟼m1+m2=b−2h and m1m2=ba}$

$\sf\pink{⇒A.M≥G.M }$

$\sf\green{ ⇒2m1+m2≥m1m2}$

$\sf\red{ ⇒(m1+m2)2≥4m1m2}$

$\sf\orange{⇒b24h2≥b4a }$

$\sf\blue{ ⇒h2≥ab}$

∴${\boxed{ h2−ab≥0}}$

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Therefore, These line passes through origin.