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

Answer 1

The correct option is D

centre and vertices

Explanation For The Correct Option:

Determining the correct statement

The Given equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1...\left(i\right)$,

Let $a=5\&b=4;a>b$

And the equation of the hyperbola is $\frac{x^2}{25}-\frac{y^2}{16}=1...\left(ii\right)$, here also a > b

Let $e\&e\text{'}$ be the eccentricities of the ellipse and hyperbola.

Therefore,

For $\left(i\right)\Rightarrow e=\sqrt{\frac{a^2–b^2}{a^2}}=\sqrt{\frac{25-16}{25}}=\frac{3}{5}$

For $\left(ii\right)\Rightarrow e\text{'}=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{25+16}{25}}=\frac{\sqrt{41}}{5}$

Thus,

The Centre of the ellipse is$(0,0)$ and the centre of the hyperbola is$(0,0)$.

Foci of the ellipse are $(\pm ae,0)$ therefore $\left(\pm 3,0\right)$ and foci of the hyperbola are $(\pm ae\prime ,0)$then $(\pm \sqrt{41},0).$

Directrices of ellipse are $x=\pm \frac{a}{e}$ $\Rightarrow x=\pm \frac{25}{3}$

Directrices of the hyperbola are $x=\pm \frac{a}{e}$ $\Rightarrow x=\pm \frac{25}{\sqrt{41}}$

Vertices of the ellipse are $(\pm a,0)$ therefore $(\pm 5,0)$ and vertices of the hyperbola are $(\pm a,0)$ therefore $(\pm 5,0)$

Hence these are common in the centre and vertices both.

Hence, option (D) is the correct answer.