The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of the directions is x = -a/e, then the equation of the normal to it at (1, 3/2) is?

x²/a² + y²/b² = 1 and e = 1/2
Directrix : x = -a/e = -a/(1/2) => a = 4
e = 4(1/2) = 2
b² = a²(1 - e²) = 4(1 - 1/4) = 4(3/4) = 3
The ellipse is : x²/4 + y²/3 = 1
Differentiating w.r.t. x on both sides, we get
2x/4 + 2yy'/3 = 0
y' = -x/4y/3 = -3x/4y
Slope of normal = 4y/3x
Slope at (1, 3/2) = 4(3/2)/3(1) = 2
y - 3/2 / x - 1 = 2 => 2y - 3 = 4(x -1) => 2y - 3 = 4x -4 => 4x - 2y = 1