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. `(pm(3sqrt5)/(2),pm(2)/(7))`
B. `(pm(3sqrt5)/(2),pm(sqrt19)/(4))`
C. `(pm2sqrt3,pm(1)/(7))`
D. `(pm2sqrt3,pm(4sqrt3)/(7))`

Answer

C

Explanation

Let P`(4costheta, 2sintheta)` be a point on the given ellipse. The equation of the normal to P is
`(4sectheta)x-(2cosectheta)y=12`
[Using : `ax sec theta -by cosec theta = a^(2)-b^(2)`]
This meets x-axis at `Q(3 costheta, 0)`
Let (h,k) be the coordinates of M i.e. the mid-point of PQ. Then,
`h=(7)/(2)costheta,ksinthetarArr((2h)/(7))^(2)+k^(2)=1`
Hence, the locus of M is `(4x^(2))/(49)+y^(2)=1`
Let e the eccentricity of the given ellipse. Then,
`e=sqrt(1-(4)/(16)=(sqrt3)/(2))`
So, the equation of the latusecta of the given ellipse are `x=pm 2sqrt3`. These lines intersect the ellipse (i) at
`(4)/(49)xx12+y^(2)=1rArry=pm(1)/(7)`
Hence, required coordinates are `(pm2sqrt3,pm(1)/(7))`