Correct Answer - D
Let the vertex A be `(a cos theta, b sin theta)`.
Since AC and AB touch the hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1`
BC is the chord of contact. Its equation is
`(x cos theta)/(a)-(y sin theta)/(b)=-1`
`"or "-(x cos theta)/(a)+(y sin theta)/(b)=1`
`"or "(x cos (pi-theta))/(a)+(y sin (pi-theta))/(b)=1`
which is the equation of the tangent to the ellipse
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
at the point `(a cos(pi-theta),b sin(pi-theta))`.
Hence, BC touches the given ellipse.