Tangents are drawn from the origin to the curve `y = sin x`. Prove that their points of contact lie on the curve `x^(2) y^(2) = (x^(2) - y^(2))`

Explanation

Let the point of contact be `(x_(1), y_(1))`
Now, `y = sin x rArr (dy)/(dx) = cos x rArr ((dy)/(dx))_((x_(1)"," y_(1))) = cos x_(1)`
`:.` the equation of the tangent at `(x_(1), y_(1))` is `(y - y_(1))/(x - x_(1)) = cos x_(1)`.
Since the tangent passes through the origin, we have `(-y_(1))/(-x_(1)) = cos x_(1)`, i.e., `(y_(1))/(x_(1)) = cos x_(1)`..(i)
But, `(x_(1) , y_(1))` lies on the curve `y = sin x`
`:. y_(1) = sin x_(1)`...(ii)
Squaring (i) and (ii) and adding, we get
`(y_(1)^(2))/(x_(1)^(2)) + y_(1)^(2) = 1 rArr y_(1)^(2) + x_(1)^(2) y_(1)^(2) = x_(1)^(2)`
`rArr x_(1)^(2) y_(1)^(2) = (x_(1)^(2) - y_(1)^(2))`
This shows that `(x_(1), y_(1))` lies on the curve `x^(2) y^(2) = (x^(2) - y^(2))`