Whakaoti i te y^2-8x-4y+4=0

$$y ^ { 2 } - 8 x - 4 y + 4 = 0$$

Whakaoti mō x

$$x=\frac{\left(y-2\right)^{2}}{8}$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te y^{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-8x-4y+4=-y^{2}$$

Me tāpiri te 4y ki ngā taha e rua.

$$-8x+4=-y^{2}+4y$$

Tangohia te 4 mai i ngā taha e rua.

$$-8x=-y^{2}+4y-4$$

Whakawehea ngā taha e rua ki te -8.

$$\frac{-8x}{-8}=-\frac{\left(y-2\right)^{2}}{-8}$$

Mā te whakawehe ki te -8 ka wetekia te whakareanga ki te -8.

$$x=-\frac{\left(y-2\right)^{2}}{-8}$$

Whakawehe -\left(y-2\right)^{2} ki te -8.

$$x=\frac{\left(y-2\right)^{2}}{8}$$

Whakaoti mō y (complex solution)

$$y=2\sqrt{2x}+2$$

$$y=-2\sqrt{2x}+2$$

Whakaoti mō y

$$y=2\sqrt{2x}+2$$

$$y=-2\sqrt{2x}+2\text{, }x\geq 0$$