Question

If an ellipse passing through (3,1)having foci(+-4,0)find 1.the length of the major axis

Answer 1

Step-by-step explanation:

Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

It passes through (3,1), so,

$\frac{9}{a^{2} } + \frac{1}{ {b}^{2} } = 1$

$\implies 9 {b}^{2} + {a}^{2} = {a}^{2} {b}^{2} \: \: \: ....(i)$

Focus $\equiv(\pm4,0)$

So,

$ae = 4$

$\implies \: a^{2} e^{2} = 16$

$\implies \: a^{2} \bigg( 1 - \frac{ {b}^{2} }{ {a}^{2} } \bigg) = 16$

$\implies \: a^{2} - {b}^{2} = 16$

$\implies \: {b}^{2} = a^{2} - 16$

Put this value in (i),

$\implies 9 ({a}^{2} - 16) + {a}^{2} = {a}^{2}( {a}^{2} - 16) \: \: \:$

$\implies 9{a}^{2} - 144 + {a}^{2} = {a}^{4} - 16 {a}^{2} \: \: \:$

$\implies 10{a}^{2} - 144 = {a}^{4} - 16 {a}^{2} \: \: \:$

$\implies {a}^{4} - 26 {a}^{2} + 144 = 0 \: \: \:$

$\implies {a}^{2} = \frac{26 \pm \sqrt{676 -576} }{2} \: \: \:$

$\implies {a}^{2} = \frac{26 \pm \sqrt{100} }{2} \: \: \:$

$\implies {a}^{2} = \frac{26 \pm 10}{2} \: \: \:$

$\implies {a}^{2} = \frac{26 + 10}{2} \: \: \:or \: \: \: {a}^{2} = \frac{26 - 10}{2}$

$\implies {a}^{2} = 18 \: \: \:or \: \: \: {a}^{2} = 8$

$\implies a = \pm 3 \sqrt{2} \: \: \:or \: \: \: {a} = \pm2 \sqrt{2}$

Taking only positive value,

$\implies a = 3 \sqrt{2} \: \: \:or \: \: \: a = 2 \sqrt{2}$

If $a=2\sqrt{2}$ then, put it in (i),

we get,

$\implies 9 {b}^{2} + {(2 \sqrt{2}) }^{2} = {(2 \sqrt{2}) }^{2} {b}^{2} \: \: \:$

$\implies 9 {b}^{2} + 8 = 8 {b}^{2} \: \: \:$

$\implies {b}^{2} + 8 = 0 \: \: \:$

$\implies {b}^{2} = - 8 \: \: \:$

Which is not possible,

So,

$\tt{ \bold{ Length \: \: of \: \: major \: \: axis = 3 \sqrt{2} }}$