Question

Q1) Determine the equation of the hyperbola which satisfies the given conditions: Foci (0, ±13), the conjugate axis is of length 24.


☞ No COPPIED and IRRELEVANT ANSWERS ❌❌​​

Answer 1

GIVEN :–

• Foci of hyperbola is (0 , ±13) .

• Length of conjugated axis is 24.

TO FIND :–

• Equation of hyperbola = ?

SOLUTION :–

▪︎ Equation of hyperbola –

$\\ \implies\bf \dfrac{ {y}^{2} }{{a}^{2}} - \dfrac{ {x}^{2} }{{b}^{2}} = 1 \: \: \: \: \: \: \: \: \: \: \: \{ \because \: \: foci \: \: on \: \: y - axis \}$

▪︎ Standard foci => (0, ±c)

• So that –

=> c = 13

• We know that –

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

• According to question –

$\\ \bf \implies Length \: \: of \: \: conjugated \: \: axis = 24$

$\\ \bf \implies 2b= 24$

$\\ \bf \implies b= \cancel\dfrac{24}{2}$

$\\ \bf \implies b=12$

• We also know that –

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

• Now put the values –

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

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

$\\ \bf \implies a= \sqrt{25}$

$\\ \bf \implies a= \pm5$

• Now put the values of 'a' & 'b' in standard equation –

$\\ \implies\bf \dfrac{ {y}^{2} }{{( \pm5)}^{2}} - \dfrac{ {x}^{2} }{{(12)}^{2}} = 1$

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

Answer 2

$\huge\sf\underline\pink{given:-}$

♡ foci (0,±13), the conjugate axis is of length 24

Here the foci are on the y-axis.

♡Therefore, the equation of the hyperbola is of the form:- 

♡$\rightarrow\frac{{y}^{2}}{{a}^{2}}-{{x}^{2}}{{b}^{2}}$

♡$\rightarrow$Since the foci are (0,±13)ae=c=13

$\small\sf\underline\pink{Since the length of the conjugate axis is}$24,⇒2b=24⇒b=12

We know that a²+b²=c²

$\rightarrow$∴a²+122=132

$\rightarrow$a²=169−144=25

Thus the equation of the hyperbola is $y²/25 -x²/144$