Question

Determine the equation of the hyperbola which satisfies the given conditions: Foci (0, ±13), the conjugate axis is of length 24.

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Answer 1

$\underline{\bf{Given:-}}$

$\green\dashrightarrow$Foci(0, +13)

$\green\dashrightarrow$Conjugate axis is of length 24.

Here we know, the Foci is on the y-axis as any point that lies on the y-axis has an x-coordinate of zero.

$\underline{\bf{To:-}}$

Determine the equation of the hyperbola which satisfies the above conditions.

$\underline{\bf{Solution:-}}$

Equation:- $\sf{\frac{y ^{2} }{ {a}^{2} } -\frac{{x}^{2}}{ {b}^{2} }}$

Now, Co-ordinates of foci = (0, ± c) & given foci = (0, ±13)

So, (0, ±c) = (0, ±13)

=> c=13

Length of conjugate axis = 2b

Length of conjugate axis = 24 (given)

=> 2b = 24

=> b=$\sf{\cancel{\frac{24}{2}}}$

=>b = 12

By Pythagoras theorem,

c²=a² + b²

Putting Values

=> (13)²= a² + (12)²

=> (13)² - (12)² = a²

=> 169 - 144 = a²

=> 25=a²

=> a² = 25

$\therefore$The required equation of ellipse,

$\sf{\frac{y ^{2} }{ {a}^{2} } -\frac{{x}^{2}}{ {b}^{2} }=1}$

Putting values,

$\sf{\frac{ {y}^{2} }{25} - \frac{ {x}^{2} }{ ({12})^{2} }=1}$

$\boxed{\bf{\frac{ {y}^{2} }{25} - \frac{ {x}^{2} }{144 }=1}}$

Answer 2

GIVEN:

• Foci of hyperbola is (0, ±13)

• Length of conjugated Axis is 24

TO FIND:

• Equation of hyperbola= ?

SOLUTION:

• Equation of hyperbola -

$\bf \implies{\frac{ {y}^{2} }{ {a}^{2} } - \frac{ {x}^{2} }{ {a}^{2} } = 1 }\: \: \: \: {\therefore foci \: on \: y - axis}$

• Standard foci => (0, + C)

• So that-

=> c = 13

• We know that-

$\bf \implies length \: of \: conjugated \: axis = 2b$

According to Question:

$\bf \implies length \: of \: conjuated \: axis \: = 24$

$\bf \: 2b = 24$

$\bf \: b = \frac{24}{2}$

$\bf b = \cancel \frac{24}{2}$

$\bf \: b = 12$

We also know that=

$\bf \: a = \sqrt{ {c}^{2} - {b}^{2} }$

• Now put the value

$\bf \: a = \sqrt{( {13})^{2} - ( {12})^{2} }$

$\bf \: a = \sqrt{169 - 144}$

$\bf a = \sqrt{25}$

$\bf \: a=±5$

Now put the value of 'a' and 'b' in standard equation-

$\bf \: \frac{ {y}^{2} }{ {(±5)}^{2} } - \frac{ {x}^{2} }{( {12}^{2}) } = 1$

${\boxed{ \bf { \implies{ \frac{ {y}^{2} }{25} - \frac{ {x}^{2} }{144} = 1}}}}$