Let `x^2+3y^2=3` be the equation of an ellipse in the `x-y` plane. `Aa n dB` are two points whose position vectors are `-sqrt(3) hat ia n d-sqrt(3) hat i+2 hat kdot` Then the position vector of a point `P` on the ellipse such that `/_A P B=pi//4` is a. `+- hat j` b. `+-( hat i+ hat j)` c. `+- hat i` d. none of these

Options

A. `pm hatj`
B. `pm (hati + hatj )`
C. `pm hati `
D. none of these

Answer

A

Explanation

Point P lies of `x^(2) + 3y^(2) = 3 " "`(i)
Now from the diagram, according to the given conditions , AP =AB
or `(x + sqrt3)^(2) + ( y-0)^(2) = 4 or (x+ sqrt3)^(2) + y^(2) =4 ` (ii)
Solving (i) and (ii), we get `x =0 and y = pm 1`
Hence, point P has position vector `pm hatj`