The foci are at: $$S(ae,0), \quad S'(-ae,0)$$ and the end of the minor axis is: $$B(0,b), \quad \text{where } b^2 = a^2(1-e^2).$$

In an equilateral triangle ∆BSS′: $$BS = SS'.$$

Now, $$BS = \sqrt{(ae)^2 + b^2}, \quad SS' = 2ae.$$ Hence, $$\sqrt{a^2e^2 + b^2} = 2ae.$$

But, $$b^2 = a^2(1-e^2).$$

So, $$\sqrt{a^2e^2 + a^2(1-e^2)} = 2ae,$$ $$\sqrt{a^2} = 2ae,$$ $$a = 2ae \;\;\Rightarrow\;\; e = \tfrac{1}{2}.$$

Given: Foci are (4, 3) and (12, 5), and the ellipse passes through the origin (0, 0).

Step 1: Use ellipse definition

$PF_1 = \sqrt{(0 - 4)^2 + (0 - 3)^2} 

= \sqrt{25} 

= 5$

$PF_2 = \sqrt{(0 - 12)^2 + (0 - 5)^2} 

= \sqrt{169} 

= 13$

Total distance = $5 + 13 = 18 \Rightarrow 2a = 18 \Rightarrow a = 9$

Step 2: Distance between the foci

$2c = \sqrt{(12 - 4)^2 + (5 - 3)^2} = \sqrt{64 + 4} = \sqrt{68} \Rightarrow c = \sqrt{17}$

Step 3: Find eccentricity

$e = \dfrac{c}{a} = \dfrac{\sqrt{17}}{9}$

✅ Final Answer: $\boxed{\dfrac{\sqrt{17}}{9}}$

Rule for Classifying Conics Using Discriminant

Given the equation: \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \)

Compute: \( \Delta = B^2 - 4AC \)

? Based on value of \( \Delta \):

• Ellipse: \( \Delta < 0 \) and \( A \ne C \), \( B \ne 0 \) → tilted ellipse
• Circle: \( \Delta < 0 \) and \( A = C \), \( B = 0 \)
• Parabola: \( \Delta = 0 \)
• Hyperbola: \( \Delta > 0 \)

Example:

For the equation: \( 3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0 \)

\( A = 3 \), \( B = 10 \), \( C = 11 \) →
\( \Delta = 10^2 - 4(3)(11) = 100 - 132 = -32 \)

Since \( \Delta < 0 \), it represents an ellipse.