JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Ellipse

From the previous question, $A=(3,0)$ and $B=\left(-\frac{9}{5},\frac{8}{5}\right)$, and $P=(3,4)$.

For an ellipse, the tangents from an external point are perpendicular to the radii at the points of tangency. The orthocenter $H$ of triangle $PAB$ can be found using the property that the altitude from $P$ to $AB$ passes through $H$.

The line $AB$ (chord of contact) has equation $x+3y=3$.

The altitude from $P(3,4)$ to $AB$ is perpendicular to $AB$, so its slope is $3$ (negative reciprocal of $-\frac{1}{3}$).

Equation of altitude from $P$:


The altitude from $B\left(-\frac{9}{5},\frac{8}{5}\right)$ to $PA$ (where $PA$ is vertical, $x=3$) is horizontal: $y=\frac{8}{5}$.

The orthocenter is the intersection of these altitudes:



Thus, the orthocenter is $\left(\frac{11}{5},\frac{8}{5}\right)$.