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.