Solve-simultaneously-for-s-in-terms-of-a-and-b-h-2-b-k-2-s-2-i-h-2-a-2-k-2-b-2-1-ii-h-s-2-2-k-b-1-s-2-4a-2-s-2-i

Question Number 49755 by ajfour last updated on 10/Dec/18

S o l v e s i m u l t a n e o u s l y f o r s i n t e r m s

o f a a n d b .

h^{2} + {(b - k)}^{2} = s^{2} \dots . . (i)

\frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} = 1 \dots . . (i i)

{(h - \frac{s}{2})}^{2} + (k + b \sqrt{1 - \frac{s^{2}}{4 a^{2}}}) = s^{2} . . (i i i) .

Answered by mr W last updated on 10/Dec/18

(i i) \times a^{2} :

h^{2} + \frac{a^{2}}{b^{2}} k^{2} = a^{2} \dots (i v)

(i) - (i v) :

b^{2} - 2 b k + k^{2} - \frac{a^{2}}{b^{2}} k^{2} = s^{2} - a^{2}

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

\Rightarrow k = \frac{b - b \sqrt{1 - (1 - \frac{a^{2}}{b^{2}}) (1 + \frac{a^{2}}{b^{2}} - \frac{s^{2}}{b^{2}})}}{(1 - \frac{a^{2}}{b^{2}})} (o n l y “ -^{″} ?)

l e t λ = \frac{a}{b}, δ = \frac{s}{b}

\Rightarrow k = \frac{b - b \sqrt{1 - (1 - λ^{2}) (1 + λ^{2} - δ^{2})}}{(1 - λ^{2})}

\Rightarrow \frac{k}{b} = \frac{1 - \sqrt{1 - (1 - λ^{2}) (1 + λ^{2} - δ^{2})}}{(1 - λ^{2})}

(i) - (i i i) :

\frac{s}{2} (2 h - \frac{s}{2}) + (b - k + k + b \sqrt{1 - \frac{s^{2}}{4 a^{2}}}) (b - k - k + b \sqrt{1 - \frac{s^{2}}{4 a^{2}}}) = 0

\frac{s}{2} (2 h - \frac{s}{2}) + b (1 + \sqrt{1 - \frac{s^{2}}{4 a^{2}}}) (b + b \sqrt{1 - \frac{s^{2}}{4 a^{2}}} - 2 k) = 0

\frac{s}{2} (2 h - \frac{s}{2}) + b^{2} (1 + \sqrt{1 - \frac{s^{2}}{4 a^{2}}}) (1 + \sqrt{1 - \frac{s^{2}}{4 a^{2}}} - \frac{2 \pm 2 \sqrt{1 - (1 - \frac{a^{2}}{b^{2}}) (1 + \frac{a^{2}}{b^{2}} - \frac{s^{2}}{b^{2}})}}{(1 - \frac{a^{2}}{b^{2}})}) = 0

s (h - \frac{s}{4}) + b^{2} (1 + \sqrt{1 - \frac{δ^{2}}{4 λ^{2}}}) (1 + \sqrt{1 - \frac{δ^{2}}{4 λ^{2}}} - \frac{2 - 2 \sqrt{1 - (1 - λ^{2}) (1 + λ^{2} - δ^{2})}}{(1 - λ^{2})}) = 0

\frac{h}{a} = \frac{δ}{4 λ} - \frac{1}{λ δ} (1 + \sqrt{1 - \frac{δ^{2}}{4 λ^{2}}}) (1 + \sqrt{1 - \frac{δ^{2}}{4 λ^{2}}} - \frac{2 - 2 \sqrt{1 - (1 - λ^{2}) (1 + λ^{2} - δ^{2})}}{(1 - λ^{2})}) = 0

(i i) :

\Rightarrow {\frac{δ}{4 λ} - \frac{1}{λ δ} (1 + \sqrt{1 - \frac{δ^{2}}{4 λ^{2}}}) (1 + \sqrt{1 - \frac{δ^{2}}{4 λ^{2}}} - \frac{2 - 2 \sqrt{1 - (1 - λ^{2}) (1 + λ^{2} - δ^{2})}}{(1 - λ^{2})})}^{2} + {\frac{1 - \sqrt{1 - (1 - λ^{2}) (1 + λ^{2} - δ^{2})}}{(1 - λ^{2})}}^{2} = 1

s o l v e f o r δ i n t e r m s o f λ \dots . (o n l y n u m m e r i c a l l y p o s s i b l e)

Commented by ajfour last updated on 10/Dec/18

T h a n k y o u S i r, A n y b e t t e r a l t e r n a t i v e

f o r Q . 49740 ?

Commented by mr W last updated on 10/Dec/18

n o s i r! I h a v e n o b e t t e r w a y .

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