Kaycee Roche

2020-10-18

Equations of Conic Sections Systems of Non-linear Equations
Solve eeach problem systematically
1. Find all values of m so that the graph $2m{x}^{2}\text{}-\text{}16mx\text{}+\text{}m{y}^{2}\text{}+\text{}7{y}^{2}=2{m}^{2}\text{}-\text{}18m$ is a circle.

Gennenzip

Skilled2020-10-19Added 96 answers

Given equation is,
$2m{x}^{2}\text{}-\text{}16mx\text{}+\text{}m{y}^{2}\text{}+\text{}7{y}^{2}=2{m}^{2}\text{}-\text{}18m$
Which can be written as,
${x}^{2}\text{}-\text{}8x\text{}+\text{}\frac{m\text{}+\text{}7}{2n}{y}^{2}=m\text{}-\text{}9$

$\Rightarrow \text{}{x}^{2}\text{}-\text{}8x\text{}+\text{}16\text{}+\text{}\frac{m\text{}+\text{}7}{2m}(y\text{}-\text{}0{)}^{2}=m\text{}-\text{}9\text{}+\text{}16$

$\Rightarrow (x\text{}-\text{}4{)}^{2}\text{}+\text{}\frac{m\text{}+\text{}7}{2m}(y\text{}-\text{}0{)}^{2}=m\text{}+\text{}7$
Equation of circle is given as,
$(x\text{}-\text{}a{)}^{2}\text{}+\text{}(y\text{}-\text{}b{)}^{2}={r}^{2}$
If the given equation is an equation of circle, then $\frac{m\text{}+\text{}7}{2m}\text{}\text{should be equal to}1\text{}and\text{}m\text{}+\text{}7\text{}\text{}0$
So,
$\frac{m\text{}+\text{}7}{2m}=1$

$m\text{}+\text{}7=2m$

$m=7$
And, $m\text{}+\text{}7=14\text{}\text{}0,$
Therefore the graph $2\text{}m{x}^{2}\text{}-\text{}16\text{}mx\text{}+\text{}m{y}^{2}\text{}+\text{}7{y}^{2}=2{m}^{2}\text{}-\text{}18m\text{}\text{will be a circle if}\text{}m=7$

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