The auxiliary circle of family of ellipses, passes through origin and makes intercept of 8 and 6 units on the xā axis and the yā axis respectively. If eccentricity of all such family of ellipse is 12, then the locus of focus of ellipse will be
A
4x2+4y2+32x−24y+75=0
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B
4x2+4y2−32x−24y+75=0
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C
4x2+4y2−32x+24y+75=0
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D
4x2+4y2−32x−24y−75=0
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Solution
The correct option is B4x2+4y2−32x−24y+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 (h−4)2+(k−3)2=254
Hence, required locus is ⇒4x2+4y2−32x−24y+75=0