Find the equation of the hyperbola with: Vertices (±5,0) , foci (±7,0) . Vertices (0,±7) , eccentricity e=2 . Foci (0,±10) , passing through (2, 3).

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: (i)Given that vertices (±5,0), foci(±7,0) Vertex of hyperbola=(±a,0) and foci(±ae,0)∴ a=5 and ae=7 ⇒5*e=7 ⇒e=75 Now b2=a2(e2−1)⇒ b2=25(4925−1) ⇒b2=25*2425 ⇒b2=24 The equation of the hyperbola is x225−y224=1(ii)Given that vertices (0,±7), e=43 Clearly, the hyperbola is vertical.∴ a=5 and e=43 We know that b2=a2(e2−1)⇒ b2=49(169−1) ⇒b2=49*79 ⇒b2=3439 The equation of the hyperbola is y249−9x2343=1 ⇒9x2−7y2+343=0(iii)Given that: foci(0,±10) ∴ ae=10 ⇒a2e2=10 We know that b2=a2(e2−1)⇒ b2=a2e2−a2 ⇒b2=10−a2 Equation of the hyperbola is y2a2−x2b2=1 ⇒y2a2−x210−a2=1 If it passes through the (2,3) then⇒9a2−410−a2=1 ⇒90−9a2−4a2a2(10−a2)=1⇒ 90−13a2=10a2−a4 ⇒a4−23a2+90=0⇒ a4−18a2−5a2+90=0⇒a2(a2−18)−5(a2−18)=0⇒ (a2−18)(a2−5)=0⇒ a2=18,a2=5∴ b2=10−18=−8 and b2=10−5=5∴ b≠−8 and b2=5Hence, the required equation is y25−x25=1 or y2−x2=5.

Find the equation of the hyperbola with:

Vertices $(\pm 5, 0)$ , foci $(\pm 7, 0)$ .

Vertices $(0, \pm 7)$ , eccentricity $e = \sqrt[]{2}$ .

Foci $(0, \pm 10)$ , passing through (2, 3).

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Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (i) G i v e n t h a t v e r t i c e s (\pm 5, 0), f o c i (\pm 7, 0) \\ V e r t e x o f h y p e r b o l a = (\pm a, 0) a n d f o c i (\pm a e, 0) \\ ∴ a = 5 a n d a e = 7 \Rightarrow 5 * e = 7 \Rightarrow e = \frac{7}{5} \\ N o w b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 25 (\frac{49}{25} - 1) \Rightarrow b^{2} = 25 * \frac{24}{25} \Rightarrow b^{2} = 24 \\ T h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{x^{2}}{25} - \frac{y^{2}}{24} = 1 \\ (i i) G i v e n t h a t v e r t i c e s (0, \pm 7), e = \frac{4}{3} \\ C l e a r l y, t h e h y p e r b o l a i s v e r t i c a l . \\ ∴ a = 5 a n d e = \frac{4}{3} \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = 49 (\frac{16}{9} - 1) \Rightarrow b^{2} = 49 * \frac{7}{9} \Rightarrow b^{2} = \frac{343}{9} \\ T h e e q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{y^{2}}{49} - \frac{9 x^{2}}{343} = 1 \Rightarrow 9 x^{2} - 7 y^{2} + 343 = 0 \\ (i i i) G i v e n t h a t : f o c i (0, \pm \sqrt[]{10}) \\ ∴ a e = \sqrt[]{10} \Rightarrow a^{2} e^{2} = 10 \\ W e k n o w t h a t b^{2} = a^{2} (e^{2} - 1) \\ \Rightarrow b^{2} = a^{2} e^{2} - a^{2} \Rightarrow b^{2} = 10 - a^{2} \\ E q u a t i o n o f t h e h y p e r b o l a i s \\ \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \Rightarrow \frac{y^{2}}{a^{2}} - \frac{x^{2}}{10 - a^{2}} = 1 \\ I f i t p a s s e s t h r o u g h t h e (2, 3) t h e n \\ \Rightarrow \frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1 \Rightarrow \frac{90 - 9 a^{2} - 4 a^{2}}{a^{2} (10 - a^{2})} = 1 \\ \Rightarrow 90 - 13 a^{2} = 10 a^{2} - a^{4} \Rightarrow a^{4} - 23 a^{2} + 90 = 0 \\ \Rightarrow a^{4} - 18 a^{2} - 5 a^{2} + 90 = 0 \\ \Rightarrow a^{2} (a^{2} - 18) - 5 (a^{2} - 18) = 0 \\ \Rightarrow (a^{2} - 18) (a^{2} - 5) = 0 \\ \Rightarrow a^{2} = 18, a^{2} = 5 \\ ∴ b^{2} = 10 - 18 = - 8 a n d b^{2} = 10 - 5 = 5 \\ ∴ b \neq - 8 a n d b^{2} = 5 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{y^{2}}{5} - \frac{x^{2}}{5} = 1 o r y^{2} - x^{2} = 5 . \end{array}$

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