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?
2026-04-24 04:28:09.1777004889
Problem in conics question
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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 $[\frac{1}{2},1]$.