Question-50638

Question Number 50638 by ajfour last updated on 18/Dec/18

Commented by ajfour last updated on 18/Dec/18

D e t e r m i n e \frac{a}{b} f o r e l l i p s e i f α = 45 ° .

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

e q n n o r m a l y = x t a n α

e q n e l l i p s e

\frac{{(x - 0)}^{2}}{a^{2}} + \frac{{(y - b)}^{2}}{b^{2}} = 1

a c o s θ, b + b s i n θ l i e s o n n o r m a l

s o

b + b s i n θ = a c o s θ t a n α

b^{2} x^{2} + a^{2} (y^{2} - 2 y b + b^{2}) = a^{2} b^{2}

b^{2} x^{2} + a^{2} y^{2} - 2 a^{2} b y + a^{2} b^{2} = a^{2} b^{2}

e q n t a n g e n t

b^{2} x (a c o s θ) + a^{2} y (b + b s i n θ) - a^{2} b (y + b + b s i n θ) = 0

x ({a b}^{2} c o s θ) + y (a^{2} b + a^{2} b s i n θ - a^{2} b) = a^{2} b (b + b s i n θ)

s l o p e = m = \frac{- {a b}^{2} c o s θ}{a^{2} b s i n θ} = - \frac{b}{a} c o t θ \leftarrow s l o p e o f t a n g e n t

s l o p e o f n o r m a l = \frac{a}{b c o t θ}

t a n α = \frac{a}{b c o t θ}

l = \sqrt{{(a c o s θ)}^{2} + {(b + b s i n θ)}^{2}} = \sqrt{a^{2} {c o s}^{2} θ + b^{2} + 2 b^{2} s i n θ + b^{2} {s i n}^{2} θ}

l c o s α = a c o s θ

l s i n α = b + b i n θ

t a n α = \frac{b + b s i n θ}{a c o s θ} = \frac{a}{b c o t θ}

\frac{b + b s i n θ}{a} = \frac{a s i n θ}{b}

a^{2} s i n θ = b^{2} + b^{2} s i n θ

s i n θ = \frac{b^{2}}{a^{2} - b^{2}}

c o t θ = \frac{b a s e}{p e r p e n d i c u l a r} = \frac{\sqrt{{(a^{2} - b^{2})}^{2} - b^{4}}}{b^{2}}

c o t θ = \frac{\sqrt{a^{4} - 2 a^{2} b^{2}}}{b^{2}}

n o w t a n α = \frac{a}{b c o t θ} = \frac{a}{b} \times \frac{b^{2}}{\sqrt{a^{4} - 2 a^{2} b^{2}}}

t a n α = \frac{a}{b} \times \frac{b^{2}}{a} \times \frac{1}{\sqrt{a^{2} - 2 b^{2}}} = \frac{b}{b \sqrt{{(\frac{a}{b})}^{2} - 2}}

t a n α = \frac{1}{\sqrt{{(\frac{a}{b})}^{2} - 2}}

g i v e n α = \frac{π}{4}

\sqrt{{(\frac{a}{b})}^{2} - 2} = 1

{(\frac{a}{b})}^{2} = 3 s o \frac{a}{b} = \sqrt{3} p l s c h e c k \dots

Commented by MJS last updated on 18/Dec/18

I get b = a \frac{\sqrt{3}}{3} \Rightarrow \frac{a}{b} = \sqrt{3}

Commented by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18

t h a n k y o u s i r i g o t t h e s a m e r e s u l t \dots

Answered by mr W last updated on 18/Dec/18

P (h, k)

\Rightarrow k = h

\frac{h^{2}}{a^{2}} + \frac{{(h - b)}^{2}}{b^{2}} = 1

\frac{h}{a^{2}} + \frac{h - b}{b^{2}} (- 1) = 0

\frac{h^{2}}{a^{2}} = \frac{(h - b) h}{b^{2}}

\frac{(h - b) h}{b^{2}} + \frac{{(h - b)}^{2}}{b^{2}} = 1

\Rightarrow h = \frac{3 b}{2}

\frac{9 b^{2}}{4 a^{2}} + \frac{b^{2}}{4 b^{2}} = 1

\frac{b^{2}}{a^{2}} = \frac{1}{3}

\Rightarrow \frac{b}{a} = \frac{1}{\sqrt{3}} \approx 0.577

Commented by mr W last updated on 18/Dec/18

\frac{x^{2}}{a^{2}} + \frac{{(y - b)}^{2}}{b^{2}} = 1

\frac{x}{a^{2}} + \frac{(y - b)}{b^{2}} y^{'} = 0

a t (h, k) : y^{'} = - 1

\Rightarrow \frac{h}{a^{2}} + \frac{(k - b)}{b^{2}} (- 1) = 0

Commented by ajfour last updated on 18/Dec/18

S i r p l e a s e e x p l a i n 4 t h l i n e

\frac{h}{a^{2}} + \frac{(h - b)}{b^{2}} (- 1) = 0

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