Question

Give me a complete and managed answer.​

Answer 1

Solution given:

The given equation of a line is

ax²+2hxy+by²=0

Let y=mx be any one line of

ax²+2hxy+by²=0

Let the perpendicular line of

y=mx is

x+my=0

According to the question the line x+my=0 is one line of Ax²+2Hxy +By²=0

Substituting value of y

Ax²+2Hx
`( - x)/(m)`+
`B (x²)/(m²)`=0

Ax²-
`( 2H)/(m)x²`+
`B (x²)/(m²)`=0

Taking LCM

we get

Ax²m²-2mHx²+Bx²=0

x²[Am²-2Hm+B]=0..............[1]

Again.

ax²+2hx(mx)+B(mx)²=0

ax²+2hmx²+Bm²x²=0

x²[a+2hm+Bm²]=0

bm²+2hn+a=0.....................[2]

Taking coefficient of equation 1 &2.

equation 1. b 2h a b 2h

equation 2.A. -2h B A -2H

Doing criss cross multiplication

ignore first coefficient and repeat first and second

again

lines are perpendicular so

`(m²)/(2hB)`=
`(m)/(aA-bB)`=
`(1)/(-2Hb-2hA)`

Taking 1st & 2nd ratio,we get,

m=
`(2hB+2Ha)/(aA-bB)`....[*]

Taking 3rd & 2nd ratio,we get,

m=
`(aA-bB)/(-2Hb-2hA)` ....[#]

Again

Equating equation * &# we get;

`(2hB+2Ha)/(aA-bB)`=
`(aA-bB)/(-2Hb-2hA)`

(aA-bB)²=-4(hB+Ha)(Hb+HA)

(aA-bB)²+4(hB+Ha)(Hb+HA)=0 is a required condition.