Question

For a positive real number $k$ let ${E_k}$ be the ellipse with equation
$\dfrac{{{x^2}}}{{{a^2} + k}} + \dfrac{{{y^2}}}{{{b^2} + k}} = 1$ .
where $a > b > 0$ .All members of the family of ellipse \left\\{ {{E_k}:k > 0} \right\\} have the same
A.Foci
B.Eccentricity
C.Pair of directrices
D.Centre

Answer 1

Hint : Here we have to find which property of the family of ellipse \left\\{ {{E_k}:k > 0} \right\\} is same when a positive real number $k$ is changes. Then understand the concepts of foci and eccentricity, pair of directions and centre of the ellipse in general form. Then check whether each option is changing or not when the values of $k$ change in the given family of ellipses.

Complete step by step solution:
Consider the standard form of the ellipse equation is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ ---(1). Where x-intercept is $a$ and y-intercept is $b$ . Whose centre is at a point $\left( {0,0} \right)$ . The formula generally associated with the focus of an ellipse is ${c^2} = {a^2} - {b^2}$ . Where $c$ is the distance from the focus to centre, $a$ is the distance from the centre to a co-vertex and $b$ is the distance from the centre to a vertex(y-axis).
The eccentricity of an ellipse (1) is $e = {\left( {1 - \dfrac{{{b^2}}}{{{a^2}}}} \right)^{\dfrac{1}{2}}}$ . The equations of the directories of the ellipse equation are $\pm \dfrac{a}{e}$ when $a > b$ .

Option (A):
Give $\dfrac{{{x^2}}}{{{a^2} + k}} + \dfrac{{{y^2}}}{{{b^2} + k}} = 1$ ---(2).
Then ${c^2} = {a^2} + k - ({b^2} + k)$ $\Rightarrow$ ${c^2} = {a^2} - {b^2}$ $\Rightarrow$ $c = \pm \sqrt {{a^2} - {b^2}}$
Hence, the foci of an ellipse independent of $k$ . Option (A) is correct.

Option (B): For the equation (2)
$e = {\left( {1 - \dfrac{{{b^2} + k}}{{{a^2} + k}}} \right)^{\dfrac{1}{2}}} = {\left( {\dfrac{{{a^2} + k - {b^2} - k}}{{{a^2} + k}}} \right)^{\dfrac{1}{2}}} = {\left( {\dfrac{{{a^2} - {b^2}}}{{{a^2} + k}}} \right)^{\dfrac{1}{2}}}$ .
$\Rightarrow$ Option (B) is incorrect.

Option (C):
Equation of the directrices of the equation (2) are $x = \pm \dfrac{{{{\left( {{a^2} + k} \right)}^{\dfrac{1}{2}}}}}{e} = \pm \dfrac{{{a^2} + k}}{{{{\left( {{a^2} - {b^2}} \right)}^{\dfrac{1}{2}}}}}$ .
Hence, the pair of directrices is not the same for all values of $k$ . Option (C) is incorrect.

Option (D): Since from the standard equation of an ellipse it is clear that the centre remains the same for all values of $k$ , Option (D) is correct.
So, the correct answer is “Option D”.

Note : Note that the y-axis is also known as vertex and the x-axis is also known as co-vertex. Also note that the eccentricity of a circle is equal to zero.