Question

Question 13 Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12

Class X1 - Maths -Conic Sections Page 262

Answer 1

concept :- if foci (±c, 0) then, standard equation of hyperbola is x²/a² - y²/b² = 1 where c² = a²+ b² and length of latusrectum = 2b²/a .

A/C to question,
given, Foci( ±4, 0) = (±c , 0)
c = 4 ,
and Length of latusrectum = 12 = 2b²/a
12 = 2b²/a
6a = b² ------------------(1)

now, use from concept ,
c² = a² + b²
here, c = 4
4² = a² + b²
16 = a² + b² -------------------(2)

from equation (1) and (2)
16 = a² + 6a
a² + 16a - 16 = 0
a² + 8a - 2a - 16 = 0
(a + 8)(a - 2) = 0
a = 2, a ≠ -8 { because a can't be negative because a is distance }
b² = 6a = 6 × 2 = 12 [ from equation (1)

hence, equation of hyperbola is
x²/4 - y²/12 = 1

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Eqn\:of\:hyperbola=3x^{2}-y^{2}=12}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = (\pm4,0)} \\ \\ \tt{ : \implies length\:of\:latus\:rectum= 12} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\\bold{As\:we\:know\:that}\\ \tt{:\implies foci=\sqrt{(4-(-4))^{2}+(0-0)^{2}}}\\\\ \tt{:\implies 2ae=8}\\\\ \tt{: \implies ae = 4}$

$\text{Both \: side \: squaring} \\ \tt{: \implies {a}^{2} {e}^{2} = 16} - - - - - (2) \\ \\ \bold{As\:we\:know\:that} \\ \tt{: \implies Length \: of \: latus \: rectum = 12 }\\ \\ \tt{: \implies \frac{2 {b}^{2} }{a} = 12} \\ \\ \tt{: \implies {b}^{2} = 6a - - - - - (3)} \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {b}^{2} = {a}^{2} {e}^{2} - {a}^{2} } \\ \\ \tt{: \implies {b}^{2} + {a}^{2} = {a}^{2} {e}^{2} } -----(4)\\ \\ \text{Putting \: values \: of \: (1) \: and \: (2) \: in \: (4)} \\ \tt{: \implies6a + {a}^{2} = 16} \\ \\ \tt{: \implies {a}^{2} + 6a - 16 = 0} \\ \\ \tt{: \implies(a - 2)(a + 8) = 0} \\ \\ \tt{: \implies a = 2 \:and \: - 8} \\ \\ \tt{: \implies a = 2} \: \: \: (Value \: of \: a \: not \: be \: negative) \\ \\ \bold{Putting \: value \: of \: a \: in \: (3)} \\ \tt{: \implies {b}^{2} = 6\times 2 } \\ \\ \green{\tt{ : \implies {b}^{2} = 12}}\\ \\ \text{Putting \: value \: of \: a \: and \: b \: in \: (1)} \\ \tt{:\implies \frac{x^{2}}{4}-\frac{y^{2}}{12}=1 }\\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: 3{x}^{2} - {y}^{2} = 12}}$