If the points of intersections of the ellipse [x^2 16 + y^2 b^2 = 1] and the circle x2 + y…

Question

If the points of intersections of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the
circle x2 + y2 = 4b, b > 4 lie on the curve y2 = 3x2, then b is equal to :

Accepted Answer

$${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ ... (1)

${x^2} + {y^2} = 4b$ .... (2)

${y^2} = 3{x^2}$ .... (3)

From eq (2) and (3)

x2 = b and y2 = 3b

From equation (1)

$${b \over {16}} + {{3b} \over {{b^2}}} = 1$$

$$ \Rightarrow {b^2} + 48 = 16b$$

$$ \Rightarrow b = 12$$

Solutions · 1