Question

$\huge \color{cyan} \: Question \: -$
Find the equations of the hyperbola satisfying the given conditions. foci (0,±√10), passing through (2,3).​

Answer 1

Answer:

Given: volume of block V=25cm3

Volume immersed in water v=20cm3

(i) If density of wood is ρgcm-3 then by principle of floatation.

Weight of block of wood = Weight of water displaced by the immersed part of block.

Vρg=v×1×g(since density of water =1gcm-3)

$\huge\red{➳Ṧřîⅉꫝᾇñ ࿐}$

Answer 2

Given, foci of hyperbola (0, ±√10) and (2,3) is a point of coordinate that passes through hyperbola

We need to find out the equation of hyperbola

Foci is on y - axis, so the structure of equation :

• y²/a² - x²/b² = 1

Now, as we know that,

→ a² + b² = c²

→ a² + b² = 10 (°•° foci = (0, ±√10))

→ a² = 10 - b²

• Let's form the required equation

$\implies\sf \dfrac{ {3}^{2} }{ {a}^{2} } - \dfrac{ {2}^{2} }{ {b}^{2} } = 1 \\ \\ \\ \implies \sf \dfrac{ 9 }{ {a}^{2} } - \dfrac{ 4 }{ {b}^{2} } = 1 \qquad( {a}^{2} = 10 - {b}^{2} ) \\ \\ \\ \implies \sf \dfrac{ 9 }{10 - {b}^{2} } - \dfrac{ 4 }{ {b}^{2} } = 1 \\ \\ \\ \implies \sf \dfrac{ 9 {b}^{2} - 4(10 - {b}^{2}) }{ {b}^{2} (10 - {b}^{2} )} = 1 \\ \\ \\ \implies \sf \dfrac{ 9 {b}^{2} - 40 + 4{b}^{2} }{ {b}^{2} (10 - {b}^{2} )} = 1 \\ \\ \\ \implies \sf 13 {b}^{2} - 40 = {b}^{2} (10 - {b}^{2} ) \\ \\ \\ \implies \sf 13 {b}^{2} - 40 = (10{b}^{2} - {b}^{4} ) \\ \\ \\ \implies \sf {b}^{4} - 40 = 10{b}^{2} - 13{b}^{2} \\ \\ \\ \implies \sf {b}^{4} - 40 = - 3 {b}^{2} \\ \\ \\ \implies \sf {b}^{4} - 40 + 3 {b}^{2} = 0 \\ \\ \\ {\footnotesize {\sf{ \pmb{ \underline{Spilt \: middle \: term}}}}} \\ \\ \\ \implies \sf {b}^{4} + 8 {b}^{2} -5 {b}^{2} - 40 = 0 \\ \\ \\ \implies \sf {b}^{2} ( {b}^{2} + 8)-5( {b}^{2} + 8)= 0 \\ \\ \\ \implies \sf ({b}^{2} - 5) ( {b}^{2} + 8)= 0 \\ \\ \\ \implies \sf b^2 = 5\:or \: - 8 \\ \\ \\ {\footnotesize {\sf{ \pmb{ \underline{Take \: {b}^{2} = 5\: and\: ignore \: -8}}}}}$

→ a² = 10 - b²

→ a² = 10 - 5

→ a² = 5

Put the value of a² and b² in the required equation

→ y²/a² - x²/b² = 1

→ y²/5 - x²/5 = 1

• Required equation : y²/5 - x²/5 = 1