Question

The equation of the pair of straight lines through origin, each of which makes as angle $\alpha$ with the line y = x, is

• A. $x^2+2xysec2\alpha +y^2=0$
• B. $x^2+2xycosec2\alpha +y^2=0$
• C. $x^2-2xycosec2\alpha +y^2=0$
• D. $x^2-2xysec2\alpha +y^2=0$

Answer 1

The correct option is D

$x^2-2xysec2\alpha +y^2=0$

Any line through the origin is y = mx.If it makes an angle $\alpha$ with the line y = x, then we should have

$tan\alpha =\pm \left\{\frac{m_1-m_2}{1+m_1{m}_2}\right\}=\pm \frac{(m-1)}{1+m}$

or$(1+m{)}^{2}ta{n}^2\alpha =(m-1{)}^2$

$\Rightarrow m^2-2m\left\{\frac{1+ta{n}^2\alpha }{1-ta{n}^2\alpha }\right\}+1=0$

$\Rightarrow m^2-2msec2\alpha +1=0$, $\because \{\frac{1+ta{n}^2\alpha }{1-ta{n}^2\alpha }=sec2\alpha \}$

But $m=\frac{y}{x}$, hence on eliminating m, we get the required equation $y^2-2xysec2\alpha +x^2=0$