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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` |
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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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