Locus of the point whose sum of distances from the origin and the x− axis is 4 units is
A
x2=8(2−y),y≥0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x2=8(2+y),y≥0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2=8(2+y),y<0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
x2=8(2−y),y<0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct options are Ax2=8(2−y),y≥0 Cx2=8(2+y),y<0 Let P(x,y) be a point which satisfies the given condition √x2+y2+|y|=4 when y≥0 then √x2+y2=4−y ⇒x2+y2=16−8y+y2 ⇒x2=8(2−y) when y<0 √x2+y2−y=4 ⇒√x2+y2=4+y ⇒x2+y2=16+8y+y2 ⇒x2=8(2+y)