A hyperbola passes through the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is

$?$ ae = 2b

$\frac{4b^2}{a^2}=e^2$  

Or 4 (1 – e²) = e²

4 = 5e² -> $e=\frac{2}{\sqrt[]{5}}$

If two circles intersect at two distinct points

->|r₁ – r₂| < C₁C₂ < r₁ + r₂

| r – 2| <  $\sqrt[]{9+16}$ < r + 2

|r – 2| < 5                     and r + 2 > 5

–5 < r – 2 < 5                    r > 3                      … (2)

–3 < r < 7                                                        … (1)

From (1) and (2)

3 < r < 7

x² – y² cosec²q = 5 $\frac{x^2}{1}-\frac{y^2}{si{n}^2\theta }=5$                        

x² cosec²q + y² = 5  $\frac{x^2}{si{n}^2\theta }+\frac{y^2}{1}=5$        

$e_H=\sqrt[]{7}{e}_e$                  

and $e_H=\sqrt[]{1+\frac{si{n}^2\theta }{1}}$ &n

Slope of axis = $\frac{1}{2}$

$y-3=\frac{1}{2}\left(x-2\right)$              

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

            α + 1.6 = 4 ⇒ α = 2.4

            β + 2.8 = 6 ⇒

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

K = $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e2 – 1)

$e=\sqrt[]{4}=2$