Question

Find the two seperate equations when the lines represented by kx^2 + 8xy - 3y^2 = 0 are perpendicular to each other

Answer 1

$\large\underline{\sf{Solution-}}$

We know that

The joint equation of line given by ax² +2hxy + by² = 0 represents two perpendicular lines iff a + b = 0.

i.e. coefficient of x² + coefficient of y² = 0.

Given equation of line is

$\rm :\longmapsto\: {kx}^{2} + 8xy - {3y}^{2} = 0$

Since, it represents two perpendicular lines.

It means, coefficient of x² + coefficient of y² = 0.

$\rm :\longmapsto\:k - 3 = 0$

$\bf\implies \:k = 3$

Now, Given equation of lines can be rewritten as

$\rm :\longmapsto\: {3x}^{2} + 8xy - {3y}^{2} = 0$

$\rm :\longmapsto\: {3x}^{2} + 9xy - xy - {3y}^{2} = 0$

$\rm :\longmapsto\:3x(x + 3y) - y(x + 3y) = 0$

$\rm :\longmapsto\:(x + 3y)(3x - y) = 0$

$\bf\implies \:x + 3y = 0 \: \: and \: \: 3x - y = 0$

Additional Information :-

The joint equation of line y = mx and y = nx is given by ax² +2hxy + by² = 0

where,

$\boxed{ \rm \: m + n = - \: \dfrac{2h}{b}}$

$\boxed{ \rm \: m n = \: \dfrac{a}{b}}$

Angle between two lines is

$\rm :\longmapsto\:tan \theta \: = \: |\dfrac{2 \sqrt{ {h}^{2} - ab} }{a + b} |$

Answer 2

Step-by-step explanation:

given,

the given equation is

kx^2 + 8xy - 3y^2 = 0............(1)

comparing with ax^2 + 2hxy + by^2 = 0

a = k, h = 4 and b = -3

now the perpendicular conditions a+b = 0

then,

k - 3 = 0

;. k = 3

now , the value of k put in equation (1) we get

3x^2 + 8xy - 3y^2 = 0 .......... (2)

so, the two seperate of equations (2)

3x^2 + 9xy - xy - 3y^2 = 0

3x( x + 3y) - y (x + 3y) =0

;. (x + 3y) (3x-y) = 0

Either,

x+3y = 0

and 3x - y =0

which is the two seperate of equations are x+3y = 0

and 3x - y =0.