The eccentricity of an ellipse whose centre is at the origin is `1/2dot` if one of its directrices is `x=-4,` then the equation of the normal to it at `(1,3/2)` is:

Options

A. `2y-x=2`
B. `4x-2y =1`
C. `4x+2y=7`
D. `x+2y=4`

Answer

B

Explanation

it is given that the centre of the ellipse is at the origin and one directix is `x=-4` So, the axes are along the coordinate axes and hence let the equation of the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` where `b^(2) = a^(2)(1-e^(2))` it is given that `x=-4` is a directrix and its eccentricity is `(1)/(2)`
` therefore (a) /(e) =4 and e=(1)/(2)implies a=2`
`Now,b^(2)=a^(2)(a-e^(2))implies b^(2)=4(1-(1)/(4))=3`
so the equation of the ellipse is
`(x^(2))/(4)+(y^(2))/(3)=1`
the equation of the normal at `(1,3//2)` is
`(a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2) or , 4x-2y=4-3 or , 4x-2y=1`