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Question

The auxiliary circle of family of ellipses, passes through origin and makes intercept of 8 and 6 units on the xāˆ’ axis and the yāˆ’ axis respectively. If eccentricity of all such family of ellipse is 12, then the locus of focus of ellipse will be

A
4x2+4y2+32x24y+75=0
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B
4x2+4y232x24y+75=0
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C
4x2+4y232x+24y+75=0
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D
4x2+4y232x24y75=0
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Solution

The correct option is B 4x2+4y232x24y+75=0


Centre of family of ellipse is (0+82,6+02)=(4,3)

Length of major axis 2a=82+62
a=5

Distance of focus from centre =ae
=5×12=52

If (h,k) is the coordinates of focus, then
(h4)2+(k3)2=254
Hence, required locus is
4x2+4y232x24y+75=0

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