orthogonally at ( then × − ax by cx = − . That is, acx bdy = , together with ( c x d y = gives, a ac = b d bd That is, = d Hence, = d Remark In the above example, the converse is also true. That is assuming the condition = d one can easily establish that the curves cut orthogonally. - - Example .
Prove that the ellipse x and the hyperbola x intersect orthogonally. Let the point of intersection of the two curves be ( , ) a b . Hence, = and a ... ( ) It is enough to show that the product of the slopes of the two curves evaluated at ( , ) a b is − .
Differentiation of x with respect x , gives y dy = Therefore dy dx = − x Then, dy a b at ( , ) = m = − Differentiation of x with respect to x , gives y dy = Therefore, dy dx = x Then dy a b at ( , ) = m Therefore, × = − × = − ... ( ) Applying the ratio of proportions in ( ), we get a b −+ = −− Therefore a . Substituting in ( ), we get m × = − . Hence, the curves cut orthogonally.