Question

The lengths of the axes of the conic 25x^2+ 4y^2-10x+4y+1 =0 are​

Answer 1

SOLUTION

TO DETERMINE

The length of the axes of the conic

$\displaystyle\sf{25 {x}^{2} + 4 {y}^{2} - 10x + 4y + 1 = 0 }$

EVALUATION

Here the given equation of the conic is

$\displaystyle\sf{25 {x}^{2} + 4 {y}^{2} - 10x + 4y + 1 = 0 }$

Which can be rewritten as below

$\displaystyle\sf{25 {x}^{2} + 4 {y}^{2} - 10x + 4y + 1 = 0 }$

$\displaystyle\sf{ \implies \: 25 {x}^{2} - 10x+ 4 {y}^{2} + 4y + 1 = 0 }$

$\displaystyle\sf{ \implies \: 25 {x}^{2} - 10x + 1+ 4 {y}^{2} + 4y + 1 = 1 }$

$\displaystyle\sf{ \implies \: {(5x - 1)}^{2} + {(2y + 1)}^{2} = 1 }$

$\displaystyle\sf{ \implies \: 25{ \bigg(x - \frac{1}{5} \bigg)}^{2} + 4{ \bigg(y + \frac{1}{2} \bigg)}^{2} = 1 }$

$\displaystyle\sf{ \implies \: \frac{{ \bigg(x - \frac{1}{5} \bigg)}^{2}}{ \frac{1}{25} } + \frac{{ \bigg(y + \frac{1}{2} \bigg)}^{2}}{ \frac{1}{4} } = 1 }$

$\displaystyle\sf{ \implies \: \frac{{ \bigg(x - \frac{1}{5} \bigg)}^{2}}{ \frac{1}{ {5}^{2} } } + \frac{{ \bigg(y + \frac{1}{2} \bigg)}^{2}}{ \frac{1}{{2}^{2} } } = 1 }$

Hence the length of the axes

$\displaystyle\sf{ = 2 \times \frac{1}{5} \: \: unit \: \: \: and \: \: 2 \times \frac{1}{2} \: \: unit }$

$\displaystyle\sf{ = \frac{2}{5} \: \: unit \: \: \: and \: \: \: 1 \: \: unit }$

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