The sides `A Ca n dA B` of a ` A B C` touch the conjugate hyperbola of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` . If the vertex `A` lies on the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` , then the side `B C` must touch (a)parabola (b) circle (c)hyperbola (d) ellipse

Options

A. parabola
B. circle
C. hyperbola
D. ellipse

Answer

D

Explanation

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.