If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies:

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

-> $\sqrt[]{1+si{n}^2\theta }=\sqrt[]{7}\sqrt[]{1-si{n}^2\theta }$

1 + sin²q = 7 – 7 sin²q

->8sin²q = 6

-> $sin\theta =\sqrt[]{\frac{3}{4}}=\frac{\sqrt[]{3}}{2}$  

-> $\theta =\frac{\pi }{3}$  

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)

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