Question

A tangent to a hyperbola `x^2/a^2 - y^2/b^2 = 1` intercepts a length of unity from each of the coordinate axes, then the point `(a, b)` lies on rectangular hyperbola

Answer 1

let the eqn of hyperbola be
`H : x^2/a^2 -y^2/b^2 = 1`
let the point on it be `(a sec theta, b tan theta)`
the equation of tangent will be
`bxsec theta - ay tan theta = ab`
putting y=0
`bxsec theta = ab `
`x= a cos theta = 1`
putting x=0
`-aytan thea = ab`
so, `y = bcos theta=1`
as `acos theta = 1`
`a = sec theta`
as `-bcos theta = 1`
`b = -tan theta`
as we know the identity `sec^2 theta - tan^2 theta = 1`
`a^2 - (-b)^2 =1`
`a^2-b^2=1 `
`x^2 - y^2=1 `