Ellipse and Functions
Consider an ellipse with its major and minor axes along the X and Y axis respectively.The ellipse has its foci at (1,0) and (-1,0) respectively and has its eccentricity as 0.5.
Let line L be a tangent to the ellipse at some point on it.Let point S be the reflection of the focus ( lying on positive X axis) in the line L. Find the locus of S.
Now consider a function f having its domain as the set of real numbers and its range a subset of the set of real numbers,this function f has the property that it is simultaneously even as well as odd function.
Calculate the area bounded between the function f and the locus of point S.It is given that the bounded area contains points whose y coordinate is non negative.
Round off the answer (ie the bounded area) to the nearest integer and submit.
The answer is 25.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Let the point of contact of tangent L and ellipse be point P. Let the foci be denoted by points M(-1,0) and N(1,0).Let point O be the origin.Now P is the midpoint of SN and O is the midpoint of MN.Therefore by midpoint theorem in triangle MNS we get MS=2OP. We know that foot of perpendicular from focus on any tangent to ellipse lies on its auxiliary circle.Therefore OP=semi major axis=2. Therefore MS=4.Hence locus of S is a circle with centre (-1,0) and radius 4.Now function f is odd as well as even function.Therefore f(-x)=f(x)=-f(x).This gives f(x)=0.Now we can easily calculate the required bounded area which comes out to be 8pi.