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klasskru [66]
3 years ago
12

Please answer the question correctly while showing detailed working. ​

Mathematics
1 answer:
balu736 [363]3 years ago
4 0

\\ \sf\longmapsto \[4x-\overline{3x-5y}-3\{2x-(3x-\overline{2x-3y})\}\]

\\ \sf\longmapsto \[4x-3x+5y-3\{2x-(3x-2x+3y)\}\]

\\ \sf\longmapsto \[x+5y-3\{2x-(x+3y)\}\]

\\ \sf\longmapsto \[x+5y-3\{2x-x-3y\}\]

\\ \sf\longmapsto \[x+5y-3(x-3y)\]

\\ \sf\longmapsto \[x+5y-3x+9y\]

\\ \sf\longmapsto x-3x+5y+9y

\\ \sf\longmapsto -2x+14y

\\ \sf\longmapsto 14y-2x

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e follSOLUTION

Given the question in the image, the following are the solution steps to answer the question

STEP 1: Write the general equation of an ellipse

\frac{\mleft(x-h\mright)^2}{a^2}+\frac{(y-h)^2}{b^2^{}}=1

STEP 2: Identify the parameters

the length of the major axis is 2a

the length of the minor axis is 2b

\begin{gathered} 2a=24,a=\frac{24}{2}=12 \\ 2b=20,b=\frac{20}{2}=10 \end{gathered}

STEP 3: Get the equation of the ellipse

\begin{gathered} By\text{ substitution,} \\ \frac{(x-h)^2}{a^2}+\frac{(y-h)^2}{b^2}=1 \\ \frac{(x-0)^2}{12^2}+\frac{(y-0)^2}{10^2}=1=\frac{x^2}{144}+\frac{y^2}{100}=1 \end{gathered}

STEP 4: Pick the nearest equation from the options,

Hence, the equation of the ellipse in the image is given as:

\frac{x^2}{144}+\frac{y^2}{95}=1

OPTION A

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