Distinguishing between Conics from General Equation and Eccentricity
The locus of ...
Question
The locus of a point (to the right of x=2) whose sum of the distances from the origin and the line x=2 is 4 units, is
A
y2=−12(x−3)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
y2=12(x−3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2=12(y−3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2=−12(y−3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Ay2=−12(x−3) Let the coordinates of the point be (h,k).
Distance of the point from origin =√(h−0)2+(k−0)2=√h2+k2
Distance of the point from the line x−2=0 is h−2
according to the given condition, √h2+k2+h−2=4⇒√h2+k2=6−h
squaring both sides, we have h2+k2=36+h2−12hork2=−12(h−3) ∴ The path of the point is y2=−12(x−3)