Let the ellipse E : x 2 144 + y 2 169 = 1 and the hyperbola H : x 2 16 - y 2 2 = - 1 have the same foci. If e and L respectively denote the eccentricity and the length of latus rectum of H, then the value of 24(e+L) is : [2026]

Question

Let the ellipse $E : \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$ and the hyperbola $H : \frac{x^{2}}{16} - \frac{y^{2}}{λ^{2}} = - 1$ have the same foci. If e and L respectively denote the eccentricity and the length of latus rectum of H, then the value of 24(e+L) is : [2026]

Smart Achievers · Accepted Answer

(3)

$Equation of hyperbola: \frac{y^{2}}{λ^{2}} - \frac{x^{2}}{16} = 1$

$Equation of ellipse: \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$

$e = \sqrt{1 - \frac{144}{169}} = \frac{5}{13}$

$Focus \Rightarrow (0, 5)$

$\Rightarrow λ \sqrt{1 + \frac{16}{λ^{2}}} = 5$

$\Rightarrow λ^{2} + 16 = 25 \Rightarrow λ = 3$

$Eccentricity of hyperbola = \sqrt{1 + \frac{16}{λ^{2}}} = \frac{5}{3}$

$Length of latus rectum of hyperbola = \frac{2 (16)}{3} = \frac{32}{3}$

$24 (e + ℓ) = 24 (\frac{5}{3} + \frac{32}{3}) = 8 \times 37 = 296$

Solution

Q.
Let the ellipse $E : \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$ and the hyperbola $H : \frac{x^{2}}{16} - \frac{y^{2}}{λ^{2}} = - 1$ have the same foci. If e and L respectively denote the eccentricity and the length of latus rectum of H, then the value of 24(e+L) is : [2026]

1 148

2 126

3 296

4 67

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