Question
Sun November 11, 2012 By:
 

An ellipse of major axis 20*3^1/2 and minor axis 20 slides along the coordinate axes and always remains confined in the 1st quadrant. The locus of the centre of the ellipse therefore describes the arc of a circle. The length of this arc is? (a) 5pi (b)20pi (c)5pi/3 (d)20pi/3

Expert Reply
Sat December 01, 2012

solution-let S(x,y)and S’(x’,y’)be the two foci of the ellipse.let (h,k) be the centre of the ellipse.

Also SS’=2ae   ,e is the eccentricity of the parabola.

(x-x’)^2+(y-y’)^2=4a^2e^2

(x+x’)^2-4xx’+(y+y’)^2-4yy’=4a^2e^2

Here x+x’=2h

         y+y’=2k

since ellipse always slides between the  coordinate axes and  remains confined in first quadrant soX axes and Y axes will be tangent two this ellipse.therefore y,y’ and x,x’ are the perpendicular distances of foci from tangents.whose product is always b^2.

So

xx’=yy’=100   a^2e^2=a^2-b^2=300-100

using all these values

h^2+k^2=400

this is the equation of circle with radius 20.

Centre will  move on an arc as describe in the next diagram,of radius 20 and angle subtended at centre will be?/6 as angle on both side will be 30 degree.

tan?=10/10?3

so ?=?/6

 

arc lenth will be 10 pi/3
Wed February 15, 2017

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