Question

x²+(y-3v2x)²=1 heeeeeeeeeellllllppppp

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Explicație pas cu pas:

1)

${x}^{2} + {(y - 3 \sqrt{2}x)}^{2} = 1 \\ {x}^{2} + {y}^{2} - 6 \sqrt{2}xy + 18 {x}^{2} - 1 = 0$

$19{x}^{2} - (6 \sqrt{2}y)x + ({y}^{2} - 1) = 0$

$\Delta = {( - 6 \sqrt{2}y)}^{2} - 4 \cdot 19 \cdot ( {y}^{2} - 1) = 72 {y}^{2} - 76 {y}^{2} - 76 = 76 - 4 {y}^{2} = 4(19 - {y}^{2})$

$\Delta \geqslant 0 \implies 19 - {y}^{2} \geqslant 0 \\ {y}^{2} \leqslant 19 \iff \boxed{- 19 \leqslant y \leqslant 19}$

$x_{1,2} = \frac{6 \sqrt{2}y \pm 2 \sqrt{19 - {y}^{2} } }{2 \cdot 19} = \\ = \frac{3 \sqrt{2}y \pm \sqrt{19 - {y}^{2} } }{19}$

2.

${(y - 3 \sqrt{2}x)}^{2} = 1 - {x}^{2}$

$1 - {x}^{2} \geqslant 0 \implies \\ {x}^{2} \leqslant 1 \iff \boxed{ - 1 \leqslant x \leqslant 1}$

$y - 3 \sqrt{2}x = \pm \sqrt{1 - {x}^{2} } \\ \iff y_{1,2} = \pm \sqrt{1 - {x}^{2} } + 3 \sqrt{2}x$