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

$\\ \color{blue} \underline{ \large \rm \: Solution :- }$

$\\ \\ \sf \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{a}{b} {x}^{2} \: + \: \frac{2hxy}{b} \: + \: {y}^{2} \: = \: 0$

$\\ \\ \sf \: \: \: \: \: : \implies \: \: \: \therefore \: {y}^{2} \: + \: \frac{2h}{b} xy \: = \: - \frac{a}{b} {x}^{2}$

$\\ \sf \: On \: completing\: the \: square \: and \: adjusting , \: we \: get$

$\\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \: \: {y}^{2} \: + \: \frac{2h}{b} xy \: + \: \frac{ {h}^{2} {x}^{2} }{ {b}^{2} } \: = \: \frac{ {h}^{2} {x}^{2} }{ {b}^{2} } \: - \: \frac{a}{b} {x}^{2}$

$\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: : \: \implies \: \: \: \bigg[ y \: + \: \frac{h}{b}x \bigg ] ^{2} \: = \: \bigg[ \frac{ ({h }^{2} \: - \: ab)}{ {b}^{2} } \bigg ]$

$\\ \\ \: \: \: \: \therefore \: \sf \: \: \ \: \: \: : \implies \: y \: + \: \frac{h}{b} x \: = \: + or -\: \frac{ \sqrt{ {h}^{2} - ab} }{b} x$

$\\ \\ \: \: \: \: \therefore \: \sf \: \: \ \: \: \: : \implies \: y \: = \: - \frac{h}{b} x \: + or - \: \frac{ \sqrt{ {h}^{2} - ab} }{b} x$

$\\ \\ \: \: \therefore \: \: \: \: \: \: : \implies \sf \: y \: = \: \bigg( \frac{ - h + or - \sqrt{ {h}^{2} - ab} }{b} \bigg)x$

$\\ \\ \sf \: \therefore \: \:\: equation \: represent \: the \: two \: lines$

$\\ \\ \: \: \therefore \: \: \: \: \: \: \sf \: y \: = \: \bigg( \frac{ - h + \sqrt{ {h}^{2} - ab} }{b} \bigg)x \: \: \: or \: \: \:y \: = \: \bigg( \frac{ - h - \sqrt{ {h}^{2} - ab} }{b} \bigg)x$

$\\ \\ \sf \: The \: above \: equation\: are \: in \: the \: form \: of \: y \: = mx \:$

These lines passing through the origin . thus the homogeneous equation (1) represent the pair of line through the origin, if h²-ab≥ 0.

Answer 2

Step-by-step explanation:

$solution$

Let two lines y=m

1

,x and y=m

2

x passess through origin.

⇒y−m

1

x=0 and y−m

2

x=0

⇒(y−m

1

x)(y−m

2

x)=0

∴y

2

−(m

1

+m

2

)xy+m

1

m

2

x

2

=0; Where m

1

and m

2

are slopes of lines.

∴y

2

−(sum of slopes)xy+(product of slopes)x

2

=0

It represents pair of straight lines passing through origin.

Now,

ax

2

+2hxy+by

2

=0

⇒y

2

+

b

2h

xy+

b

a

x

2

=0

⇒m

1

+m

2

=

b

−2h

and m

1

m

2

=

b

a

⇒A.M≥G.M

⇒

2

m

1

+m

2

≥

m

1

m

2

⇒(m

1

+m

2

)

2

≥4m

1

m

2

⇒

b

2

4h

2

≥

b

4a

⇒h

2

≥ab

∴h

2

−ab≥0

Here ur solution

Mark as brainleist please