Question

An ellipse intersects the hyperbola $3x^2-y^2=9$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

• A. the equation of the ellipse is $3x^2+4y^2=3b^2$
• B. the foci of the ellipse are $(\pm 2,0)$
• C. The equation of the ellipse is $3x^2+6y^2=16$
• D. the foci of the ellipse are $(\pm 2\sqrt{3},0)$

Answer 1

The correct option is A the equation of the ellipse is $3x^2+4y^2=3b^2$

Equation of hyperbola:

$3x^2-y^2=9$

$\Rightarrow \frac{x^2}{3}-\frac{y^2}{9}=1\Rightarrow a^2=3,b^2=9$

$e=\sqrt{\frac{a^2+b^2}{a^2}}$

$\therefore e=\sqrt{\frac{3+9}{3}}=2$

Eccentricity of parabola is reciprocal to the eccentricity of hyperbola.

$e=\frac{1}{2}$

Eccentricity of parabola :

$e=\sqrt{\frac{a^2-b^2}{a^2}}$

$\frac{1}{2}=\sqrt{\frac{a^2-b^2}{a^2}}$

$\frac{1}{4}=\frac{a^2-b^2}{a^2}$

$a^2=4a^2-4b^2$

$\Rightarrow a^2=\frac{4}{3}{b}^2$

Therefore equation of ellipse is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{\frac{4}{3}{b}^2}+\frac{y^2}{b^2}=1$

$x^2+\frac{4}{3}{y}^2=\frac{4}{3}{b}^2$

$3x^2+4y^2=3b^2$