Question

From a fixed point A on the circumference of a circle of radius r, the perpendicular AY is let fall on the tangent at P. The maximum area of the triangle APY is-

• A. r²
• B. $\frac{3\sqrt{3}}{4}$r²
• C. $\frac{3\sqrt{3}}{8}$r²
• D. $\sqrt{3}$r²

Answer 1

Answer: (C)

$\frac{3\sqrt{3}}{8}$r²

Answer 2

OP ^ PY

ŠAPY = 90 – q OPA = q

ŠPAY = q

Now DOPA

AP² = r² + r² – 2rr cos (p – 2q) = 4r²cos2q

AP = 2r cos2q PY = AP sin q = r sin2q

AY = AP cos q = 2Y cos 2q

\ Area of DAPY = 1/2 PY. AY

= r²sin2q cos2q

$\frac{d\Delta}{d\theta}$ = r² [2cos 2q cos2q – sin² 2q] = 0

q = $\frac{\pi}{2},\frac{\pi}{6}$

q ¹ $\frac{\pi}{2}$ D is maximum at q = $\frac{\pi}{6}$

D_max = r² . $\frac{\sqrt{3}}{2}$. $\left( \frac{\sqrt{3}}{2} \right)^{2}$ = $\frac{3\sqrt{3}r^{2}}{8}$