Question

Solve the following : If the curves ax² + by² = 1 and a'x² + b'y² = 1, intersect orthogonally, then prove that `(1)/a - (1)/b = (1)/a' - (1)/b'`.

Answer 1

Let P(x₁, y₁) be the point of intersection of the curves.

∴ ax₁² + by₁² = 1 and a'x₁² + b'y₁² = 1
∴ ax₁² + by₁² = a'x₁² + b'y₁²
∴ (a – a')x₁² = (b' – b)y₁²                         ...(1)
Differentiating ax² + by² = 1 w.r.t. x, we get

`a xx 2x + b xx 2y dy/dx = 0`

∴ `dy/dx = (-ax)/(by)`
∴ slope of the tangent at (x₁, y₁) = m₁

= `(dy/dx)_("at" (x_1, y_1)) = (-a'x_1)/(b'y_1)`
Since. curves intersect each other orthogonally,
m₁m₂ = – 1

∴ `((-ax_1)/(by_1))((-a'x_1)/(b'y_1))` = – 1

∴`(aa'x_1^2)/(b  b'y_1^2)` = – 1

∴ `(aa')/(b b') = (-y_1^2)/(x_1^2)`

∴ `(aa')/(b b') = ((a - a')/(b - b'))`                  ...[By (1)]

∴ `(a - a')/(aa') = (a - a')/(b b')`

∴ `(1)/a' - (1)/a = (1)/b' - (1)/b`

∴ `(1)/a - (1)/b = (1)/a' - (1)/b'`.