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An ellipse has OB as semi-minor axis, F and F’ are its foci and the angle FBF’ is a right angle then, the eccentricity of the ellipse is
a.13b.14c.12d.12

Answer
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Hint: In order to solve this problem we need to know that the Product of slopes of two perpendicular lines is -1. Drawing the diagram will help you a lot. You need to use the formula of eccentricity e=1b2a2. Doing this will solve this problem.

Complete step-by-step answer:
The figure to this problem can be drawn as,
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Let equation of ellipse is x2a2+y2b2=1
We can easily see in figure coordinates of B(0, b), F (ae, 0) and F’(-ae, 0).
We know, F’B perpendicular to FB.
Thus, Product of slopes of two perpendicular lines is -1.
So, slope of F’B x slope of FB = -1
Slope of F’B = b00+ae=bae
Slope of FB = b00ae=bae
bae×bae=1 (When we multiply the slopes of two perpendicular lines)
 b2a2e2=1
On solving we get,
b2=a2e2…..(1)
We know that e=1b2a2
Now, put the value of b2 from (1) equation in (2) equation.
We get the new equation as,
e=1a2e2a2e=1e2
On squaring both sides we get,
e2=1e2
On further solving the equations we get,
2e2=1e=12

So, the correct answer is “Option d”.

Note: Whenever we face such types of problems we use some important points. Like first of all draw a figure and mark coordinates then find the value of slope of lines using coordinates and as we know the product of slopes of two perpendicular lines always be -1. Knowing this will solve our problem and will give you the right answer.