Question

The ellipse $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{{16}} = 1$ and the hyperbola $\dfrac{{{x^2}}}{{25}} - \dfrac{{{y^2}}}{{16}} = 1$ have in common
1)centre only
2)centre, foci and directrices
3)centre, foci and vertices
4)centre and vertices

Answer 1

Hint : This question requires basic concepts of ellipse and hyperbola like eccentricities, vertices, centre, directrices. The formula involved in this question are:
The eccentricity of an ellipse $e = \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}}$
The eccentricity of a hyperbola $e = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}}$
Foci of the ellipse are $x = \pm ae$
Foci of the hyperbola are $x = \pm ae$
Vertices of an ellipse $x = \pm a,y = \pm b$
Vertices of a hyperbola $x = \pm a,y = \pm b$
Centre of an ellipse $x = 0,y = 0$
Centre of a hyperbola $x = 0,y = 0$
Directrices of an ellipse $x = \pm \dfrac{a}{e}$
Directrices of a hyperbola $x = \pm \dfrac{a}{e}$
Where $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is the ellipse, and $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is the hyperbola.

Complete step-by-step answer :
Let’s begin this question by solving the eccentricities of both the curves
For the ellipse, the eccentricity is given by
$\Rightarrow e = \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}}$
$\Rightarrow e = \sqrt {\dfrac{{25 - 16}}{{25}}}$
$\Rightarrow e = \sqrt {\dfrac{9}{{25}}}$
$\Rightarrow e = \dfrac{3}{5}$
For the hyperbola, the eccentricity is given by
$\Rightarrow e = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}}$
$\Rightarrow e = \sqrt {\dfrac{{25 + 16}}{{25}}}$
$\Rightarrow e = \sqrt {\dfrac{{41}}{{25}}}$
$\Rightarrow e = \dfrac{{\sqrt {41} }}{5}$
The eccentricities of the two curves are not equal.
Now, let’s find the centre of both the curves
For, the ellipse the centre is given by
$\Rightarrow x = 0,y = 0$
For, the hyperbola the centre is given by
$\Rightarrow x = 0,y = 0$
The centre of the two curves is equal.
Now, let’s find the foci of the two curves
For the ellipse, the foci are given by
$\Rightarrow x = \pm ae$
$\Rightarrow x = \pm 5 \times \dfrac{3}{5}$
$\Rightarrow x = \pm 3$
For the hyperbola, the foci are given by
$\Rightarrow x = \pm ae$
$\Rightarrow x = \pm 5 \times \dfrac{{\sqrt {41} }}{5}$
$\Rightarrow x = \pm \sqrt {41}$
Now, let’s find the vertices of the curve
For, the ellipse the vertices are given by
$\Rightarrow x = \pm a,y = \pm b$
$\Rightarrow x = \pm 5,y = \pm 4$
For, the hyperbola the vertices are given by
$\Rightarrow x = \pm a,y = \pm b$
$\Rightarrow x = \pm 5,y = \pm 4$
The vertices of the two curves are equal.
Now, let’s find the directrices of the curves
For, the ellipse the directrices are given by
$\Rightarrow x = \pm \dfrac{a}{e}$
$\Rightarrow x = \pm \dfrac{5}{3} \times 5$
$\Rightarrow x = \pm \dfrac{{25}}{3}$
For, the hyperbola the directrices are given by
$\Rightarrow x = \pm \dfrac{a}{e}$
$\Rightarrow x = \pm \dfrac{5}{{\sqrt {41} }} \times 5$
$\Rightarrow x = \pm \dfrac{{25}}{{\sqrt {41} }}$
The directrices of the two curves are not equal.
After solving and analyzing, the centre and vertices are equal.
Thus, option(4) is correct.
So, the correct answer is “Option 4”.

Note : This question requires the basic concepts of ellipse and hyperbola. One should be familiar with terms like centre, vertices, directrices, etc., to solve this question. Do not commit calculation mistakes, and be sure of the final answer.