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Question Number 45639 by peter frank last updated on 14/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18

	$1) \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{d}{d x} (\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = 0$ $\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \times \frac{d y}{d x} = 0$ $\frac{d y}{d x} = - \frac{\frac{2 x}{a^{2}}}{\frac{2 y}{b^{2}}} = - \frac{b^{2} x}{a^{2} y}$ $s o s l o p e o f t a n g e n t a t p o i n t (a c o s θ, b s i n θ)$ $m = - \frac{b^{2} \times a c o s θ}{a^{2} \times b s i n θ} = - \frac{b c o s θ}{a s i n θ}$ $s o e q n o f t a n g e n g i s$ $(y - b s i n θ) = - \frac{b c o s θ}{a s i n θ} (x - a c o s θ)$ $y a s i n θ - {a b s i n}^{2} θ + x b c o s θ - {a b c o s}^{2} θ = 0$ $y a s i n θ + x b c o s θ = a b$ $\frac{y s i n θ}{b} + \frac{x c o s θ}{a} = 1$ $2) \frac{x c o s θ}{a} + \frac{y s i n θ}{b} = 1$ $\frac{x}{\frac{a}{c o s θ}} + \frac{y}{\frac{b}{s i n θ}} = 1$ $c o m p a r e t h i s e q n w i t h s t r a i g h t t l i n e i n i n t e r c e p t e d$ $f o r m$ $\frac{x}{A} + \frac{y}{B} = 1 A = \frac{a}{c o s θ} = i n t e r c e p t i n x a x i s$ $B = \frac{b}{s i n θ} i n t e r c e p t o n y a x i s$ $s o a r e a o f t r i a n g l e$ $\frac{1}{2} \times \frac{a}{c o s θ} \times \frac{b}{s i n θ}$ $= \frac{a b}{s i n 2 θ}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18
	$3)A(a,0) B(0,b) P(acosθ,bsinθ) area of triangle formula (1/2)∣{x_1 (y_2 −y_3 )+x_2 (y_3 −y_1 )+x_3 (y_1 −y_2 )}∣ =(1/2)∣{a(b−bsinθ)+0(bsinθ−0)+acosθ(0−b)}∣ =(1/2)∣{ab−absinθ−abcosθ}∣$
	$3) A (a, 0) B (0, b) P (a c o s θ, b s i n θ)$ $a r e a o f t r i a n g l e f o r m u l a$ $\frac{1}{2} ∣ {x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})} ∣$ $= \frac{1}{2} ∣ {a (b - b s i n θ) + 0 (b s i n θ - 0) + a c o s θ (0 - b)} ∣$ $= \frac{1}{2} ∣ {a b - a b s i n θ - a b c o s θ} ∣$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18

	$d) a r e a o f A P B = \frac{1}{2} a b (c o s θ + s i n θ - 1)$ $a l r e a d y p r o v e d$ $S = \frac{a b}{2} (c o s θ + s i n θ - 1)$ $\frac{d S}{d θ} = \frac{a b}{2} (- s i n θ + c o s θ)$ $f o r m a x / m i n \frac{d S}{d θ} = 0 = \frac{a b}{2} (- s i n θ + c o s θ)$ $s i n θ = c o s θ s o t a n θ = 1 = t a n \frac{π}{4} θ = \frac{π}{4}$ $\frac{d^{2} S}{d θ^{2}} = \frac{a b}{2} (- s i n θ - c o s θ) = - \frac{a b}{2} (s i n θ + c o s θ)$ $i n t h e i n t e r v a l 0 < θ < \frac{π}{2}$ $s i n θ + c o s θ = + v e s o \frac{d^{2} θ}{d θ^{2}} = - v e$ $n o w p o i n t A = (a, 0) B = (0, b)$ $s l o p e o f A B m_{1} = \frac{b - 0}{0 - a} = - \frac{b}{a}$ $t a n g e n t a t p i s (a l r e a d y p r o v e d)$ $\frac{x c o s θ}{a} + \frac{y s i n θ}{b} = 1 a t θ = \frac{π}{4}$ $\frac{x}{a \sqrt{2}} + \frac{y}{b \sqrt{2}} = 1 \frac{y}{b \sqrt{2}} = 1 - \frac{x}{a \sqrt{2}}$ $y = b \sqrt{2} - \frac{b}{a} x$ $s l o p e i s m_{2} = - \frac{b}{a}$ $s o m_{1} = m_{2} = \frac{- b}{a} t h a t m e a n s A B i s p a r a l l e l$ $t o t a n g e n t a t p a t θ = \frac{π}{4}$ $p l s c h e c k . . .$
	Commented by peter frank last updated on 19/Oct/18

	$the ans w er given is \frac{1}{2} abcosec θ$
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