# Equation of a Hyperbola referred to Two Perpendicular Lines

## Equation of a Hyperbola referred to Two Perpendicular Lines

Consider the hyperbola $$\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$$.

From figure

Let P (x₁, y₁) be any point on the hyperbola.

Then, PM = y

and PN = x

Therefore, $$\frac{P{{N}^{2}}}{{{a}^{2}}}-\frac{P{{M}^{2}}}{{{b}^{2}}}=1$$.

If the perpendicular distances P₁ and P₂ of a moving point P (x, y) from two mutually perpendicular coplanar straight lines

L₁ ≡ a₁ x + b₁ y + c₁ = 0,

L₂ ≡ b₁ x – a₁ y + c₂ = 0, respectively, satisfy the equation.

$$\frac{P{{N}^{2}}}{{{a}^{2}}}-\frac{P{{M}^{2}}}{{{b}^{2}}}=1$$.

$$\frac{{{p}_{1}}^{2}}{{{a}^{2}}}-\frac{{{p}_{2}}^{2}}{{{b}^{2}}}=1$$.

$$\frac{{{\left\{ \frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}} \right\}}^{2}}}{{{a}^{2}}}-\frac{{{\left\{ \frac{{{b}_{1}}x-{{a}_{1}}y+{{c}_{2}}}{\sqrt{{{b}_{1}}^{2}+{{a}_{1}}^{2}}} \right\}}^{2}}}{{{b}^{2}}}=1$$.

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

Then the locus of point P denotes a hyperbola on the plane of the given lines such that

(i) The center of the hyperbola is the point of intersection of the line L₁ = 0 and L₂ = 0.

(ii) The transverse axis lies along L₂ = 0 and the conjugate axis lies along L₁ = 0.

(iii) The lengths of the transverse and conjugate axes are 2a and 2b, respectively, if a > b.