Find-the-shortest-distance-from-the-origin-to-the-hyperbola-x-2-8xy-7y-2-225-z-0-

Question Number 144534 by imjagoll last updated on 26/Jun/21

Find the shortest distance from

the origin to the hyperbola

x^{2} + 8 xy + {7 y}^{2} = 225, z = 0

Answered by liberty last updated on 26/Jun/21

we must find the minimum value

of x^{2} + y^{2} (the square of the distance

from the origin to any point in

the xy plane) subject to constraint

x^{2} + 8 xy + {7 y}^{2} = 225

by Langrange multiplier

\emptyset = x^{2} + 8 xy + {7 y}^{2} + λ (x^{2} + y^{2})

\emptyset_{x} = 2 x + 8 y + 2 λ x = 0 or (λ + 1) x + 4 y = 0

ϕ_{y} = 8 x + 14 y + 2 λ y = 0 or 4 x + (λ + 7) y = 0

| \begin{matrix} λ + 1 4 \\ 4 λ + 7 \end{matrix} | = 0 \to {\begin{cases} λ = - 9 \\ λ = 1 \end{cases}

when λ = - 9 we get y = 2 x and

substitution in x^{2} + 8 xy + {7 y}^{2} = 225

yields x^{2} = 5, y^{2} = 20 so the shortest

distance is \sqrt{5 + 20} = 5

when λ = 1 give x = - 2 y and

substitution in x^{2} + 8 xy + {7 y}^{2} = 225

yields - {5 y}^{2} = 225 for which no

real solution exists .

Answered by mr W last updated on 26/Jun/21

r = d i s t a n c e f r o m (0, 0) t o (x, y)

x = r \cos θ

y = r \sin θ

r^{2} \cos^{2} θ + 8 r^{2} \cos θ \sin θ + 7 r^{2} \sin^{2} θ = 225

r^{2} (4 + 4 \sin 2 θ - 3 \cos 2 θ) = 225

r^{2} = \frac{225}{4 + 5 \sin (2 θ - \tan^{- 1} \frac{3}{4})} ⩾ \frac{225}{4 + 5}

\Rightarrow r_{m i n} = \sqrt{\frac{225}{9}} = 5

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