The joint equation of pair of lines through origin each of which makes of \(60^{\circ}\) with the \(y\)-axis…

The joint equation of pair of lines through origin each of which makes of \(60^{\circ}\) with the \(y\)-axis is
  1. \(3 x^2+y^2=0\)
  2. \(x^2-3 y^2=0\)
  3. \(x^2+3 y^2=0\)
  4. \(3 x^2-y^2=0\)

Solution

A line making \(60^{\circ}\) with y axis can make angles \(30^{\circ}\) and \(150^{\circ}\) with \(\mathrm{x}\) axis. The equations can be written as
\(y=\tan \left(30^{\circ}\right) x \text { and } y=\tan \left(150^{\circ}\right) x\)
Thus, the equations become
\(\mathrm{y}=\frac{1}{\sqrt{3}} \mathrm{x} \text { and } \mathrm{y}=\frac{-1}{\sqrt{3}} \mathrm{x}\)
The combined equation can be written as
\((\sqrt{3} y-x) \cdot(\sqrt{3} y+x)=0\)
i.e. \(3 y^2-x^2=0 \text { or } x^2-3 y^2=0\)

Asked in: MHT CET 2020 (20 Oct Shift 1)

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