Question

Show that the combined equation of the pair of lines passing through the origin and each making an angle α with the line x + y = 0 is x² + 2(sec 2α)xy + y² = 0

Answer 1

Let OA and OB be the required lines.

Let OA (or OB) has slope m.

∴ its equation is y = mx        ...(1)

It makes an angle α with x + y = 0 whose slope is - 1.

∴ tan α = ${\displaystyle \left|\frac{\text{m}+1}{1+\text{m}(-1)}\right|}$

Squaring both sides, we get,

${\displaystyle {\text{tan}}^2\alpha =\frac{{(\text{m}+1)}^2}{{(1-\text{m})}^2}}$

∴ tan²α (1 - 2m + m²) = m² + 2m + 1

∴ tan²α - 2mtan²α + m²tan²α = m² + 2m + 1

∴ (tan²α - 1)m² - 2(1 + tan²α)m + (tan²α - 1) = 0

∴ ${\displaystyle {\text{m}}^2-2\left(\frac{1+{\text{tan}}^2\alpha }{{\text{tan}}^2\alpha -1}\right)\text{m}+1=0}$

∴ ${\displaystyle {\text{m}}^2+2\left(\frac{1+{\text{tan}}^2\alpha }{1-{\text{tan}}^2\alpha }\right)\text{m}+1=0}$

∴ ${\displaystyle {\text{m}}^2+2\left(\sec 2\alpha \right)\text{m}+1=0}$ ...${\displaystyle [\because \text{cos 2}\alpha =\frac{1-{\text{tan}}^2\alpha }{1+{\text{tan}}^2\alpha }]}$

∴ ${\displaystyle \frac{{\text{y}}^2}{{\text{x}}^2}+2\left(\text{sec}2\alpha \right)\frac{\text{y}}{\text{x}}+1=0}$

∴ ${\displaystyle {\text{y}}^2 2\text{xy} \text{sec}2\alpha +{\text{x}}^2=0}$   ...[By (1)]

∴ ${\displaystyle {\text{y}}^2+2\text{xy}\text{sec 2}\alpha +{\text{x}}^2=0}$

∴ x² + 2(sec 2α)xy + y² = 0 is the required equation.