# Find the length of axes of the conic $9{x^2} + 4{y^2} - 6x + 4y + 1 = 0$

$

A.{\text{ }}\dfrac{1}{2},9 \\

B.{\text{ 3,}}\dfrac{2}{5} \\

C.{\text{ }}1,\dfrac{2}{3} \\

D.{\text{ 3,2}} \\

$

Last updated date: 23rd Mar 2023

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Answer

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**Hint:**In order to solve these types of question, firstly we have to convert the given equation into the equation of an ellipse by simplifying the equation into $\dfrac{{{{\left( {x - a} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y + b} \right)}^2}}}{{{b^2}}}$ = 0 such that we will get the value of $a$ and $b$ to get the length of major and minor axes and we will get our desired answer.

**Complete step-by-step answer:**

We have given that,

$9{x^2} + 4{y^2} - 6x + 4y + 1 = 0$

$\Rightarrow$ ${\left( {3x} \right)^2} - 6x + {\left( {2y} \right)^2} + 4y + 1 = 0$

$\Rightarrow$ ${\left( {3x} \right)^2} - 6x + {\left( {2y + 1} \right)^2} = 0$

Adding and subtracting square of $\left( { + 1, - 1} \right)$ on R.H.S, we get

$\Rightarrow$ ${\left( {3x} \right)^2} - 6x + {1^2} - {1^2} + {\left( {2y + 1} \right)^2} = 0$

$\Rightarrow$ ${\left( {3x - 1} \right)^2} - 1 + {\left( {2y + 1} \right)^2} = 0$

$\Rightarrow$ ${\left( {3x - 1} \right)^2} + {\left( {2y + 1} \right)^2} = 1$

$\Rightarrow$ ${\left( {3\left( {x - \dfrac{1}{3}} \right)} \right)^2} + {\left( {2\left( {y + \dfrac{1}{2}} \right)} \right)^2} = 1$

$\Rightarrow$ $9{\left( {x - \dfrac{1}{3}} \right)^2} + 4{\left( {y + \dfrac{1}{2}} \right)^2} = 1$

Converting this into $\dfrac{{{{\left( {x - a} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y + b} \right)}^2}}}{{{b^2}}} - - - - - - \left( 1 \right)$,

Thus

${\dfrac{{\left( {x - \dfrac{1}{3}} \right)}}{{\dfrac{1}{9}}}^2} + {\dfrac{{\left( {y + \dfrac{1}{2}} \right)}}{{\dfrac{1}{4}}}^2} = 1$

$\Rightarrow$ ${\dfrac{{\left( {x - \dfrac{1}{3}} \right)}}{{{{\left( {\dfrac{1}{3}} \right)}^2}}}^2} + {\dfrac{{\left( {y + \dfrac{1}{2}} \right)}}{{{{\left( {\dfrac{1}{2}} \right)}^2}}}^2} = 1 - - - - - \left( 2 \right)$

Hence this is the equation of ellipse.

Comparing equation (1) and (2), we get

Therefore,$a = \dfrac{1}{3}$ and $b = \dfrac{1}{2}$

Thus, $b > a$

Length of the major axes $ = 2b = 2 \times \dfrac{1}{2} = 1$

Length of minor axes $ = 2a = 2\dfrac{1}{3} = \dfrac{2}{3}$

**Note:**Whenever we face such a type of question the key concept is that we have to convert the given equation in the form of an ellipse equation such that we will easily find out the value of $a,b$ by comparing the equations .

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