**Equation of an Ellipse Referred to Two Perpendicular Lines**

Consider the ellipse \(\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\).

Let P (x, y) be any point on the ellipse.

Then PM = y and PN = x.

\(\frac{P{{N}^{2}}}{{{a}^{2}}}+\frac{P{{M}^{2}}}{{{b}^{2}}}=1\).

If follows from this if the perpendicular distance P₁ and P₂ of a moving point P (x, y) from two mutually perpendicular coplanar straight lines L₁ ≡ a₁x + b₁y + c₁ = 0, L₂ ≡ b₁x – a₁y + c₂ = 0 respectively satisfy the equation.

\(\frac{{{P}^{2}}}{{{a}^{2}}}+\frac{{{P}^{2}}}{{{b}^{2}}}=1\).

\(\frac{\{({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}})/\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}\}}{{{a}^{2}}}+\frac{\{({{b}_{1}}x+{{a}_{1}}y+{{c}_{1}})/\sqrt{{{b}_{1}}^{2}+{{a}_{1}}^{2}}\}}{{{b}^{2}}}=1\).

Then the locus of point P is an ellipse in the plane of given lines such that,

1. The center of the ellipse is the point of intersection of the lines L₁ = 0 and L₂ = 0.

2. The major axis lies along L₂ = 0 and the minor axis lies along L₁ = 0, if a > b.

**Example**: Find the equation of the ellipse whose axes are of length 6 and 2√6 and their equations are x – 3y + 3 = 0 and 3x + y – 1 = 0, respectively.

**Solution: **Given that,

x – 3y + 3 = 0 and 3x + y – 1 = 0.

The equation of the ellipse is \(\frac{\{({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}})/\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}\}}{{{a}^{2}}}+\frac{\{({{b}_{1}}x+{{a}_{1}}y+{{c}_{1}})/\sqrt{{{b}_{1}}^{2}+{{a}_{1}}^{2}}\}}{{{b}^{2}}}=1\).

\(\frac{\{(x-3y+3)/\sqrt{{{1}^{2}}+9}\}}{\sqrt{6}}+\frac{\{(3x+y-{{1}_{1}})/\sqrt{9+{{1}^{2}}}\}}{6}=1\).

\(\frac{{{(x-3y+3)}^{2}}}{60}+\frac{{{(3x+y-1)}^{2}}}{90}=1\).

3 (x – 3y + 3)² + 2 (3x + y – 1)² = 180

21x² – 6xy + 20y² + 6x – 58y – 151 = 0.