Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity …

Question

Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is $${1 \over 2}$$. If P(1, $$\beta $$), $$\beta $$ > 0 is a point on this ellipse, then the equation of the normal to it at P is :

Accepted Answer

$$e = {1 \over 2}$$

$$x = {a \over e} = 4$$

$$ \Rightarrow $$ a = 2

$${e^2} = 1 - {{{b^2}} \over {{a^2}}} $$

$$\Rightarrow {1 \over 4} = 1 - {{{b^2}} \over 4}$$

$${{{b^2}} \over 4} = {3 \over 4} \Rightarrow {b^2} = 3$$

$$ \therefore $$ Ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$

P(1, $$\beta $$) is on the ellipse

$${1 \over 4} + {{{\beta ^2}} \over 3} = 1$$

$${{{\beta ^2}} \over 3} = {3 \over 4} \Rightarrow \beta  = {3 \over 2}$$

$$ \Rightarrow P\left( {1,{3 \over 2}} \right)$$

Equation of normal $${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$

$$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$

$$ \Rightarrow $$ $4x - 2y = 1$

Solutions · 1