Phrases Previous Year Questions (PYQs) – Ellipse – Page 1 of 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

$e={\displaystyle \frac{c}{a}}={\displaystyle \frac{\sqrt{17}}{9}}$

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

Rule for Classifying Conics Using Discriminant

Given the equation: $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Compute: $\mathrm{\Delta }=B^2-4AC$

? Based on value of $\mathrm{\Delta }$:

• Ellipse: $\mathrm{\Delta }<0$ and $A\ne C$, $B\ne 0$ → tilted ellipse
• Circle: $\mathrm{\Delta }<0$ and $A=C$, $B=0$
• Parabola: $\mathrm{\Delta }=0$
• Hyperbola: $\mathrm{\Delta }>0$

Example:

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

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

Since $\mathrm{\Delta }<0$, it represents an ellipse.