The normal to the ellipse $E : \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ at the point $P (2 \cos ϕ, \sqrt{3} \sin ϕ)$ meets the x-axis at Q and y-axis at R. Let C be its centre
Then which of the following is / are true

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If $\tan ϕ = \frac{\sqrt{3}}{2}, (0 < ϕ < \frac{π}{2})$ then area of triangle CRQ is $\frac{1}{14} s q$

If $\tan ϕ = \frac{\sqrt{3}}{2}, (0 < ϕ < \frac{π}{2})$ then area of triangle CRQ is $\frac{1}{7} s q$

Perpendicular distance of centre of ellipse E, from the normal, drawn to it at P can be 1

perpendicular distance of centre of ellipse E, and normal drawn to it at P can be 1.5

answer is A, C.

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Detailed Solution

A normal of slope 1 of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is drawn. It meets the x-axis & the y-axis in P&Q respectively. If C is the centre of the ellipse, show that the area of triangle CPQ is $\frac{{(a^{2} - b^{2})}^{2}}{2 (a^{2} + b^{2})} s q$
Max distance between centre of ellipse and any normal drawn to it is $|a - b|$ when the distance is max $\tan^{2} θ = \frac{b}{a}$