# Problem in conics question

A vertical line passing through the point ($h$,0) intersects the ellipse $$\frac{x^2}{4}+\frac{y^2}{3}=1$$ at the points P & Q.Let the tangents to ellipse at P & Q meet at the point R.If $\Delta(h)$=area of the triangle PQR.How can i find the maximum and minimum value of $\Delta(h)$ for $\frac{1}{2} \le h \le 1$. My try:Vertical line passes through ($h$,0).The equation of line is $x=h$.[General point$(h,y)$] It intersects the ellipse so it should satisfy the ellipse and on solving $$\frac{h^2}{4}+\frac{y^2}{3}=1.$$Now what can i do next?

Solve for $y$. You will get two points. Then find the third one as an intersection of tangents. Then there is a formula for surface of a triangle when all three vertices are known. This will of course be a function of $h$ and then you have to find a maximum of one-variable function on $[\frac{1}{2},1]$.
• Ok.i got P as ($h,3\mathrm{ \sqrt{1-h^2/4}}$)and Q as ($h,-3\mathrm{ \sqrt{1-h^2/4}}$).But how to find the intersecting point? – Devgeet Patel Jan 31 '14 at 16:55
• You mean $R$? You have to find the tangents through these $P$ and $Q$. Do you know how to do this? – Poppy Jan 31 '14 at 16:56
• Yeah i should take the equation $y=mx \pm \mathrm{\sqrt{a^2m^2+b^2}}$.Is that correct? – Devgeet Patel Jan 31 '14 at 16:59