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Ellipse question

2005 · Shift 0 · Q107

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Ellipse
2005 · Shift 0 · Q107

Ellipse question

2005 · Shift 0 · Q107

JEE MainMathematicsEllipseMCQ+4 / −1

An ellipse has

OB

as semi minor axis,

F

and

F

' its focii and theangle

FBF

' is a right angle. Then the eccentricity of the ellipse is :

A
${1 \over {\sqrt 2 }}$
B
${1 \over 2}$
C
${1 \over 4}$
D
${1 \over {\sqrt 3 }}$

View written solutionFree

Correct answer: A

Set up the standard ellipse

Take the ellipse centered at $O(0,0)$ with:

semi-major axis $= a$
semi-minor axis $= b$
foci at $F(c,0)$ and $F'(-c,0)$

Then, $c^2=a^2-b^2, \qquad e=\frac{c}{a}.$

Since $OB$ is the semi-minor axis, the endpoint $B$ of the minor axis is $B(0,b).$

Use the condition $\angle FBF' = 90^\circ$

Vectors: $\overrightarrow{BF}=(c,-b), \qquad \overrightarrow{BF'}=(-c,-b).$

If the angle between them is $90^\circ$ , then their dot product is zero: $\overrightarrow{BF}\cdot \overrightarrow{BF'}=0.$

So, $(c)(-c)+(-b)(-b)=0$ $-c^2+b^2=0$ $b^2=c^2.$

Thus, $b=c.$

Relate $a,b,c$

Using $c^2=a^2-b^2,$ and $c^2=b^2$ , we get $b^2=a^2-b^2$ $a^2=2b^2.$

Hence, $a=\sqrt{2}\,b, \qquad c=b.$

Find the eccentricity

$e=\frac{c}{a}=\frac{b}{\sqrt{2}b}=\frac{1}{\sqrt{2}}.$

Check options

The correct option is $\boxed{\frac{1}{\sqrt{2}}}$ which is Option A.

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