Transform the equation $y^{2} + 4 y cot α - 4 x = 0$ from rectangular axes to oblique axes meeting at an angle $α$ , the axis of $x$ being kept the same.

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Solution

To change one set of coordinate axes inclined at an angle $ω$ to another system at $ω^{'}$ and origin remain unchanged we substitute
$x sin ω = x^{'} sin (ω - θ) + y^{'} sin (ω - ω^{'} - θ) y sin ω = x^{'} sin θ + y^{'} sin (ω^{'} + θ)$
$ω = 90^{\circ}, ω^{'} = α$ and $θ = 0^{\circ}$
$\Rightarrow x sin 90^{\circ} = x^{'} sin (90^{\circ} - 0^{\circ}) + y^{'} sin (90^{\circ} - α - 0^{\circ}) \Rightarrow x = x^{'} + cos α y^{'} . . . . . . . (i) \Rightarrow y sin 90^{\circ} = x^{'} sin 0^{\circ} + y^{'} sin (α + 0^{\circ}) \Rightarrow y = y^{'} sin α . . . . . . . . . (i i)$
Given equation is $y^{2} + 4 y cot α - 4 x = 0$
Substituting $(i)$ and $(i i)$ we get ,
${(y^{'} sin α)}^{2} + 4 (y^{'} sin α) cot α - 4 (x^{'} + cos α y^{'}) = 0 {y^{'}}^{2} {sin}^{2} α + 4 y^{'} cos α - 4 x^{'} - 4 y^{'} cos α = 0 {y^{'}}^{2} = 4 x^{'} {csc}^{2} α$ $

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