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Chapter 15: Problem 1411

The centre of the ellipse \(\left[(\mathrm{x}+\mathrm{y}-2)^{2} / 9\right]+\left[(\mathrm{x}-\mathrm{y})^{2} / 16\right]=1\) is \(\ldots\) (a) \((1,1)\) (b) \((0,0)\) (c) \((0,1)\) (d) \((1,0)\)

Short Answer

Expert verified
The center of the ellipse is (d) \((1,0)\).

Step by step solution

01

Rewrite the given equation into suitable form

We start by expanding the given equation: \[\begin{aligned} \left[\frac{(x+y-2)^{2}}{9}\right]+\left[\frac{(x-y)^{2}}{16}\right] &=1 \\ \frac{x^2+2xy-4x+y^2-4y+4}{9} + \frac{x^2 -2xy +y^2}{16} &=1 \end{aligned}\]
02

Combine like terms

Now, we will combine like terms (i.e., collect and sum up all the variables together): \[\frac{(16 + 9)(x^2)-36y+9y^2}{144} =1\]
03

Rewrite the equation into standard form

We will now rewrite the equation into the standard form of an ellipse: \[\frac{x^2}{9} - \frac{2y}{1} + \frac{y^2}{16} = 1\] Comparing the last equation to the standard equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), we can conclude that \(x-h=0 \Rightarrow x=h\) and \(y-k=0 \Rightarrow y=k\).
04

Identify the center (h, k)

Now we have the ellipse equation in the form: \[\frac{x^2}{9} + \frac{(y-2)^2}{16} = 1\] Comparing the equation to the standard form, we can see that (h, k) = (1,0) which corresponds to the answer choice: (d) \((1,0)\)

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Most popular questions from this chapter

One of the diameters of the circle circumscribing the rectangle \(\mathrm{ABCD}\) is \(\mathrm{x}-4 \mathrm{y}+7=0 .\) If \(\mathrm{A}\) and \(\mathrm{B}\) are points \((-3,4)\) and \((5,4)\) respectively, then the area of the rectangle is ... (a) 32 sq. units (b) 16 sq. units (c) 64 sq. units (d) 8 sq. units

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