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Eccentricity of Ellipse Foci (3,3),(-4,4) Through Origin | JEE

JEE Maths question with a full step-by-step solution.

Question
An ellipse having foci at (3,3)(3,3) and (−4,4)(-4,4) and passing through the origin has eccentricity equal to:
A37\dfrac{3}{7}
B27\dfrac{2}{7}
C57\dfrac{5}{7}correct
D35\dfrac{3}{5}
Solution
Let S1=(3,3)S_1=(3,3), S2=(−4,4)S_2=(-4,4) be the foci and P=(0,0)P=(0,0) a point on the ellipse. Step 1: For any point on an ellipse, the sum of focal radii equals 2a2a.
PS1=32+32=18=32,PS2=(−4)2+42=32=42.PS_1=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2,\qquad PS_2=\sqrt{(-4)^2+4^2}=\sqrt{32}=4\sqrt2.
⇒2a=PS1+PS2=32+42=72.\Rightarrow 2a=PS_1+PS_2=3\sqrt2+4\sqrt2=7\sqrt2.
 Step 2: The distance between the foci equals 2ae2ae.
S1S2=(3+4)2+(3−4)2=49+1=50=52  ⇒  2ae=52.S_1S_2=\sqrt{(3+4)^2+(3-4)^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt2 \;\Rightarrow\; 2ae=5\sqrt2.
 Step 3: Divide.
e=2ae2a=5272=57.e=\frac{2ae}{2a}=\frac{5\sqrt2}{7\sqrt2}=\frac57.
 Correct answer: (3)
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