# Find the equation of the hyperbola with foci (0,±10), passing through (2, 3) - Mathematics

Sum

Find the equation of the hyperbola with foci (0, +- sqrt(10)), passing through (2, 3)

#### Solution

Given that: foci (0, +- sqrt(10))

∴ ae = sqrt(10)

⇒ a^2e^2 = 10

We know that b^2 = a^2(e^2 - 1)

⇒ b^2 = a^2e^2 - a^2

⇒ b^2 = 10 - a^2

Equation of hyperbola is y^2/a^2 - x^2/b^2 = 1

⇒ y^2/a^2 - x^2/(10 - a^2) = 1

If it passes through the point (2, 3) then

9/a^2 - 4/(10 - a^2) = 1

⇒ (90 - 9a^2 - 4a^2)/(a^2(10 - a^2)) = 1

⇒ 90 – 13a2 = a2(10 – a2)

⇒ 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

∴ b2 = 5

Here, the required equation is y^2/5 - x^2/5 = 1 or y2 – x2 = 5.

Concept: Hyperbola - Standard Equation of Hyperbola
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 11 Conic Sections
Exercise | Q 32.(c) | Page 204