Question

The centre of ellipse $\dfrac{{{{\left( {x + y - 2} \right)}^2}}}{9} + \dfrac{{{{\left( {x - y} \right)}^2}}}{{16}} = 1$ is
A.$\left( {0,0} \right)$
B.$\left( {1,0} \right)$
C.$\left( {0,1} \right)$
D.$\left( {1,1} \right)$

Answer 1

Hint : In this problem , we have to find the centre of ellipse . We will compare the given equation of ellipse with the standard equation $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ and then form equation to find the values of $x$ and $y$ . Hence the centre of the ellipse will be $\left( {x,y} \right)$.

Complete step-by-step answer :
We are given the equation of ellipse is
$\dfrac{{{{\left( {x + y - 2} \right)}^2}}}{9} + \dfrac{{{{\left( {x - y} \right)}^2}}}{{16}} = 1$ .
Comparing with the standard equation of the ellipse, i.e.
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ , where $0 < b < a$.
We have
$x + y - 2 = 0$ ……………………………….(1)
$x - y = 0$ …………………………….(2)
From the equation (2) , we have $x = y$.
Putting the value of $x = y$, from the equation (2) in equation (1), we can get
$\Rightarrow 2y - 2 = 0$

$$\Rightarrow 2y = 2 \\\ \Rightarrow y = 1 \;$$

Hence from equation (2), we get $x = y = 1$.
Therefore, The centre of the ellipse is $\left( {x,y} \right) = \left( {1,1} \right)$.
Hence option D is the correct answer.
So, the correct answer is “Option D”.

Note : Ellipse is the path traced by a point which moves in a plane in such a way that the sum of its distance between two fixed points in the plane is constant. The two fixed points are called the focus of the ellipse.