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.