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
•
Total views: 306k
•
Views today: 3.83k
Answer
306k+ views
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 .
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 .
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
