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Eccentricity from Foci (-2,4),(2,1) and Minor Axis End | JEE

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

Question
The foci of an ellipse are (−2,4)(-2,4) and (2,1)(2,1). The point (1,236)\left(1,\dfrac{23}{6}\right) is an extremity of the minor axis. What is the value of the eccentricity?
A913\dfrac{9}{13}
B313\dfrac{3}{\sqrt{13}}correct
C213\dfrac{2}{\sqrt{13}}
D32\dfrac{3}{2}
Solution
Step 1: The centre is the midpoint of the foci.
C=(−2+22,4+12)=(0,52).C=\left(\frac{-2+2}{2},\frac{4+1}{2}\right)=\left(0,\frac52\right).
 Step 2: The distance between the foci is 2ae2ae.
2ae=(2+2)2+(1−4)2=16+9=5  ⇒  ae=52.2ae=\sqrt{(2+2)^2+(1-4)^2}=\sqrt{16+9}=5 \;\Rightarrow\; ae=\frac52.
 Step 3: The minor-axis extremity lies at distance bb from the centre.
b=(1−0)2+(236−52)2=1+(86)2=1+169=259=53.b=\sqrt{(1-0)^2+\left(\tfrac{23}{6}-\tfrac52\right)^2}=\sqrt{1+\left(\tfrac{8}{6}\right)^2}=\sqrt{1+\tfrac{16}{9}}=\sqrt{\tfrac{25}{9}}=\frac53.
 Step 4: Use b2=a2−a2e2b^2=a^2-a^2e^2 to find a2a^2.
259=a2−254  ⇒  a2=259+254=25(4+9)36=32536.\frac{25}{9}=a^2-\frac{25}{4}\;\Rightarrow\; a^2=\frac{25}{9}+\frac{25}{4}=\frac{25(4+9)}{36}=\frac{325}{36}.
 Step 5: Then
e2=a2e2a2=25/4325/36=225325=913  ⇒  e=313.e^2=\frac{a^2e^2}{a^2}=\frac{25/4}{325/36}=\frac{225}{325}=\frac{9}{13}\;\Rightarrow\; e=\frac{3}{\sqrt{13}}.
 Correct answer: (2)
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