The normal at a point `P` on the ellipse `x^2+4y^2=16` meets the x-axis at `Qdot` If `M` is the midpoint of the line segment `P Q ,` then the locus of `M` intersects the latus rectums of the given ellipse at points. `(+-((3sqrt(5)))/2+-2/7)` (b) `(+-((3sqrt(5)))/2+-(sqrt(19))/7)` `(+-2sqrt(3),+-1/7)` (d) `(+-2sqrt(3)+-(4sqrt(3))/7)`

Options

A. `(+-3sqrt(5)//2,+-2//7)`
B. `(+-3sqrt(5)//2,+-sqrt(19)//7)`
C. `+-2sqrt(3),+-17)`
D. `(+-2sqrt(3),+-4sqrt(3)//7)`

Explanation

Given ellipse is `(x^(2))/(16)+(y^(2))/(4)=1`. Normal at `P(4 cos phi, 2 sin phi)`,
given by `4x sec phi-2y "cosec"phi=12`
It meets x-axist at
Let `M-=(alpha,beta)`
be the point of PQ.

`:. alpha=(3cos phi+4 cos phi)/(2)`
`=(7)/(2)cos phi`
or `cos phi=(2)/(7)alpha`
Using `cos^(2)phi+sin^(2)phi=1`, we have
`(4)/(49)alpha^(2)+beta^(2)=1`
or `(4)/(49)x^(2)+y^(2)=1`
Now, the rectum lines of given ellipse are `x=+-2sqrt(3)`
Solving (1) and (2), we have
`(48)/(49)+y^(2)=1`
`or y=+-(1)/(7)`
The points of intersection are `(+-2sqrt(3),+-1//7)`