CP and CD are conjugate semi-diameters of the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1`, The locus of the mid-point of PD, is

Options

A. `x^(2)/a^(2) + y^(2)/b^(2) = 2`
B. `x^(2)/a^(2) + y^(2)/b^(2) = 1/2`
C. `x^(2)/a^(2) + y^(2)/b^(2) = 4`
D. `x^(2)/a^(2) + y^(2)/b^(2) = 1/4`

Answer

B

Explanation

Let (h, k) be the mid-point of PD (See Fig. 11)
Then,
`2h = a cos theta - sin theta` and `2k = b sin theta + b cos theta`
`rArr (2h)/(a) = cos theta - sin theta` and `(2k)/(b) = sin theta + cos theta`
`rArr (4h^(2))/(a^(2)) + (4k^(2))/(b^(2)) = (cos theta - sin theta)^(2) + (sin theta + cos theta)^(2)`
`rArr h^(2)/a^(2) + k^(2)/b^(2) = 1/2`
Hence, the locus of (h, k) is `x^(2)/a^(2) + y^(2)/b^(2) = 1/2`.