Question

The ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$, have in common

• A. Centre only
• B. Centre, foci and directrices
• C. Centre, foci and vertices
• D. Centre and vertices only

Answer 1

The correct option is D Centre and vertices only

Equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1,$ $a>b$ and equation of hyperbola is $\frac{x^2}{25}-\frac{y^2}{16}=1,$ $a>b$.
Let $e$ and $e^{{}^{\prime }}$ be the eccentricities of the ellipse and hyperbola.
Therefore, $e=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{25-16}{25}}$
$=\frac{3}{5}$
and $e^{{}^{\prime }}=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{25+16}{25}}$
$=\frac{\sqrt{41}}{5}$
(i) Centre of ellipse $(0,0)$ and centre of hyperbola is $(0,0)$.
(ii) Foci of ellipse are $(\pm ae,0)$ or $(\pm 3,0)$. Foci of hyperbola are $(\pm a{e}^{{}^{\prime }},0)$ or $(\pm \sqrt{41},0)$.
(iii) Directrices of ellipse are $x=\pm \frac{a}{e}$
$\Rightarrow x=\pm \frac{25}{3}$ directrices of hyperbola are
$x=\pm \frac{a}{e}$
$\Rightarrow x=\pm \frac{25}{\sqrt{41}}$
(iv) Vertices of ellipse are $(\pm a,0)$ or $(\pm 5,0)$. Vertices of hyperbola are $(\pm a,0)$ or $(\pm 5,0)$. From the above discussions, their are common in centre and vertices.