Question

Please solve this question ​

Answer 1

Given equation is,

$\longrightarrow 100y^2-44x^2=275$

$\longrightarrow\dfrac{100y^2-44x^2}{275}=1$

$\longrightarrow\dfrac{y^2}{\left(\dfrac{275}{100}\right)}-\dfrac{x^2}{\left(\dfrac{275}{44}\right)}=1$

Now this equation is in the form $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}= 1.$

So it represents a hyperbola with transverse axis along y axis with,

• $a^2=\dfrac{275}{100}$

• $b^2=\dfrac{275}{44}$

Now,

$\longrightarrow c^2=a^2+ b^2$

$\longrightarrow c^2=\dfrac{275}{100}+\dfrac{275}{44}$

$\longrightarrow c^2=9$

$\longrightarrow c=3$

Hence the coordinates of the foci are $\mathbf{(0,\ \pm3)}.$

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• What are the coordinates of the foci of

$\tt \: {100 \: y}^{2} - 44 \: {x}^{2} = 275$

─━─━─━─━─━─━─━─━─━─━─━─━─

Solution:-

$\tt \: {100 \: y}^{2} - 44 \: {x}^{2} = 275$

☆ Divide both sides be 275, we get

$\tt \: \dfrac{100 \: {y}^{2} }{275} - \dfrac{44 \: {x}^{2} }{275} = 1$

$\tt \implies \: \dfrac{ {y}^{2} }{\dfrac{275}{100} } - \dfrac{ {x}^{2} }{\dfrac{275}{44} } = 1$

$\tt \: On \: comparing \: with \: \: \: \dfrac{ {y}^{2} }{ {b}^{2} } - \dfrac{ {x}^{2} }{ {a}^{2} } = 1$

$\tt \implies \: \: {b}^{2} = \dfrac{275}{100} \: and \: {a}^{2} = \dfrac{275}{44}$

$\tt \: We \: know, \: {c}^{2} = {a}^{2} + {b}^{2}$

$\tt \implies \: {c}^{2} = \dfrac{275}{100} + \dfrac{275}{44}$

$\tt \implies \: {c}^{2} = 275 \times \bigg(\dfrac{1}{100} + \dfrac{1}{44} \bigg)$

$\tt \implies \: {c}^{2} \: = \cancel{275} \times \bigg(\dfrac{44 + 100}{ \cancel{4400} \: \: ^{16}} \bigg)$

$\tt \implies \: {c}^{2} = \dfrac{ \cancel{144} \: \: ^9}{ \cancel{16}}$

$\tt \implies \: {c}^{2} = 9$

$\tt \implies \: c \: = \: 3 \: \:$

$\tt \:the \: coordinates \: of \: foci \: = (0, \pm \: c) = (0, \pm \: 3)$