First slide

Parabola

Question

$P$ is a point on the parabola $y^{2} = 4 a x$ whose ordinate
is equal to its abscissa and $P Q$ is a focal chord, $R$
and $S$ are the feet of the perpendiculars from $P$ and
$Q$ respectively on the tangent at the vertex, $T$ is the
foot of the perpendicular from $Q$ to $P R$ , area of the
triangle $P T Q$ is

Moderate

A

$75 a^{2} / 4$
B

$85 a^{2} / 2$
C

$75 a^{2} / 8$
D

$45 a^{2} / 2$

Solution

$P (4 a, 4 a), Q (a / 4, - a), R (0, 4 a), S (0, - a)$
$T (a / 4, 4 a)$ . so area of the $∆ P T Q$ is
$\frac{1}{2} \times P T ∣ \times Q T = \frac{1}{2} (4 a - \frac{a}{4}) \times 5 a = \frac{75}{8} a^{2}$ .

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