Tangents are drawn from the point P(3, 4) to | vedclass

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(C) The equation of the chord of contact $AB$ for the point $P(3, 4)$ with respect to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is given by $T =

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$\left( \frac{11}{5}, \frac{8}{5} \right)$

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(C) The equation of the chord of contact $AB$ for the point $P(3, 4)$ with respect to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is given by $T = 0$,which is $\frac{3x}{9} + \frac{4y}{4} = 1$,simplifying to $\frac{x}{3} + y = 1$,or $x + 3y = 3$.The points of contact are $A(3, 0)$ and $B(-9/5, 8/5)$.The orthocenter $R$ is the intersection of the altitudes. The altitude from $A$ to $PB$ is perpendicular to $PB$. The slope of $PB$ is $\frac{4 - 8/5}{3 - (-9/5)} = \frac{12/5}{24/5} = \frac{1}{2}$. Thus,the slope of the altitude from $A$ is $-2$. The equation is $y - 0 = -2(x - 3)$,or $2x + y = 6$.The altitude from $B$ to $PA$ is perpendicular to $PA$. The slope of $PA$ is $\frac{4 - 0}{3 - 3}$,which is undefined (vertical line). Thus,the altitude from $B$ is a horizontal line passing through $B(-9/5, 8/5)$,which is $y = 8/5$.Solving the system $2x + y = 6$ and $y = 8/5$ gives $2x + 8/5 = 6$,so $2x = 22/5$,which means $x = 11/5$.Therefore,the orthocenter $R$ is $\left( \frac{11}{5}, \frac{8}{5} \right)$.

Tangents are drawn from the point $P(3, 4)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,touching the ellipse at points $A$ and $B$. The orthocenter of $\Delta PAB$ is:

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