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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)

and

(2,1)

. The point

\left(1,\dfrac{23}{6}\right)

is an extremity of the minor axis. What is the value of the eccentricity?

\dfrac{9}{13}

\dfrac{3}{\sqrt{13}}

correct

\dfrac{2}{\sqrt{13}}

\dfrac{3}{2}

Solution

Step 1: The centre is the midpoint of the foci.

C=\left(\frac{-2+2}{2},\frac{4+1}{2}\right)=\left(0,\frac52\right).

Step 2: The distance between the foci is

2ae

2ae=\sqrt{(2+2)^2+(1-4)^2}=\sqrt{16+9}=5 \;\Rightarrow\; ae=\frac52.

Step 3: The minor-axis extremity lies at distance

b

from the centre.

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

b^2=a^2-a^2e^2

to find

a^2

\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

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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