Question

Find k if the slope of one of the lines given by 3x² + 4xy + ky² = 0 is three times the other.

Answer 1

3x² + 4xy + ky² = 0

∴ divide by x² 

${\displaystyle \frac{{\text{3x}}^2}{{\text{x}}^2}+\frac{\text{4xy}}{{\text{x}}^2}+\frac{{\text{ky}}^2}{{\text{x}}^2}=0}$

${\displaystyle 3+\frac{\text{4y}}{\text{x}}+\frac{{\text{ky}}^2}{{\text{x}}^2}=0}$   ....(1)

∴ y = mx

∴ ${\displaystyle \frac{\text{y}}{\text{x}}=\text{m}}$

Put ${\displaystyle \frac{\text{y}}{\text{x}}=\text{m}}$ in equation (1)

Comparing the equation km² + 4m + 3 = 0 with ax² + 2hxy + by² = 0, we get,

a = k , 2h = 4, b = 3

m₁ = 3m₂     ...(given condition)

m₁ + m₂ = ${\displaystyle \frac{\text{-2h}}{\text{k}}=-\frac{4}{\text{k}}}$

m₁m₂ = ${\displaystyle \frac{\text{a}}{\text{b}}=\frac{3}{\text{k}}}$

m₁ + m₂ = ${\displaystyle -\frac{4}{\text{k}}}$

4m₂ = ${\displaystyle -\frac{4}{\text{k}}}$     .....(m₁ = 3m₂)

m₂ = ${\displaystyle -\frac{1}{\text{k}}}$

m₁m₂ =${\displaystyle \frac{3}{\text{k}}}$

${\displaystyle 3{\text{m}}_2^2=\frac{3}{\text{k}}}$     .......(m₁ = 3m₂)

${\displaystyle 3{(-\frac{1}{\text{k}})}^2=\frac{3}{\text{k}}}$      ......(m₂ = ${\displaystyle -\frac{1}{\text{k}}}$)

${\displaystyle \frac{1}{{\text{k}}^2}=\frac{1}{\text{k}}}$

${\displaystyle {\text{k}}^2=\text{k}}$

k = 1  or k = 0