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The eccentricity of an ellipse whose centre is atthe 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:`4x+2y=7`(2) `x+2y=4`(3) `2y-x=2`(4) `4x-2y=1`

Answer» Eccentricity of the given ellipse`, e = 1/2`
Equation of the directrix,
`x = -4`.
`:. +- a/e = -4`
`=> +-a/(1/2) = -4`
`=> a = +-2`
`=>a^2 = 4`
Now, we know, `e^2 = 1- b^2/a^2`
`=>1/4 = 1-b^2/4`
`=>b^2/4 = 3/4`
`=>b^2 = 3`
Now, equation of normal to an ellipse at point `(x_1,y_1)` is given by,
`(a^2x)/(x_1) -(b^2y)/(y_1) = a^2-b^2`
So, the equation of normal to the given ellipse at point `(1,3/2)` will be,
`4x/1-3y/(3/2) = 4-3`
`=>4x-2y = 1.`


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