The orthocenter of the triangle PAB is, Tangents are drawn from the point P(3, 4) to the ellipse touching the ellipse at points A and B. [2010]

Question

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. [2010]

Q. The orthocenter of the triangle PAB is

Smart Achievers · Accepted Answer

(3)

$Let H be the orthocentre of ∆ P A B, then as B H ⊥ A P,$ $B H is a horizontal line through B .$

$∴$ $y-coordinate of B = \frac{8}{5}$

$Let H has coordinate (α, \frac{8}{5}) .$

$∴$ $Slope of P H = \frac{\frac{8}{5} - 4}{α - 3} = \frac{- 12}{5 (α - 3)}$

$and slope of A B = \frac{\frac{8}{5} - 0}{- \frac{9}{5} - 3} = \frac{8}{- 24} = - \frac{1}{3}$

$But P H ⊥ A B,$ $∴$ $\frac{- 12}{5 (α - 3)} \times (\frac{- 1}{3}) = - 1$

$\Rightarrow 4 = - 5 α + 15 or α = \frac{11}{5}$ $∴$ $H (\frac{11}{5}, \frac{8}{5}) .$

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