Question

show that pair of bisector of angle between the lines ( ax+by)²=c(bx-ay)²(c>0) are parallel and perpendicular to the line ax+by+k=0​

Answer 1

we have to show that the pair of bisector of angle between the lines (ax + by)² = c(bx - ay)² (c > 0) are parallel and perpendicular to the line ax + by + k = 0..

solution : here, (ax + by)² = c(bx - ay)²

⇒a²x² + b²y² - 2abxy = cb²x² + ca²y² + 2abcxy

⇒x²(a² - b²c) + y²(b² - a²c) + (2abc + 2ab)xy = 0

on comparing with standard equation,

we get, a = (a² - b²c) , b = (b² - a²c) , h = (abc + ab)

now equation of bisector of angle is given by,

(x² - y²)/xy = (a - b)/h

⇒(x² - y²)/xy = [(a² - b²c) - (b² - a²c)]/(abc + ab)

⇒(x² - y²)/xy = (a² - b²)/ab

⇒abx² - aby² = (a² - b²)xy

let m1 and m2 are slope of bisector of angle.

so, m1 + m2 = -2h/b = -(a² - b²)/ab = b/a - a/b

m1m2 = - 1

so, m1 = b/a and m2 = -a/b

now here slope of line ax + by + k = 0, is -a/b

here it is clear that -a/b = m1, i.e., one of bisectors of angle is parallel to line

and -a/b × m2 = -1 , so one of bisectors is perpendicular to line.

Hence proved

Answer 2

Answer:

no not getting but it is wrong