Question

If x = `a(t - 1/t)`, y = `a(t + 1/t)`, where t is the parameter, then `dy/dx` =?

Options

• `y/x`

• `(-x)/y`

• `x/y`

• `(-y)/x`

Answer 1

`bb(x/y)`

Explanation:

We have, x = `a(t - 1/t)`, y = `a(t + 1/t)`

Then, y² – x² = `[a^2(t + 1/t)^2 - a^2(t - 1/t)^2]`

`\implies` y² – x² = 4a²

 After differentiating on both sides w.r.t. 'x', we get

`2y dy/dx - 2x` = 0

`\implies 2(y  dy/dx - x)` = 0

Hence, `dy/dx = x/y`