Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 32501 by riza kesk last updated on 26/Mar/18

$${(\mathrm{x}-2\mathrm{y}+3)}^2+{(3\mathrm{x}+4\mathrm{y}-1)}^2=100$$$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{ellipse}?$$

Answered by MJS last updated on 27/Mar/18

$$\mathrm{the}\mathrm{center}\mathrm{of}$$$${ax}^2+bxy+{cy}^2+dx+ey+f=0$$$$\mathrm{is}M=\left(\begin{array}{c}\frac{be-2cd}{4ac-b^2}\\ \frac{bd-2ae}{4ac-b^2}\end{array}\right)$$$${(x-2y+3)}^2+{(3x+4y-1)}^2=100$$$$x^2+2xy+2y^2-2y-9=0$$$$M=\left(\begin{array}{c}-1\\ 1\end{array}\right)$$$$x\to x-1$$$$y\to y+1$$$$x^2+2xy+2y^2-10=0$$$$\tan 2\theta =\frac{b}{a-c}=-2$$$$\Rightarrow \tan \theta =\frac{1}{2}-\frac{\sqrt{5}}{2}\vee \tan \theta =\frac{1}{2}+\frac{\sqrt{5}}{2}$$$$\mathrm{axis}\mathit{a}:y=(\frac{1}{2}-\frac{\sqrt{5}}{2})x$$$$x^2+2(\frac{1}{2}-\frac{\sqrt{5}}{2})x^2+2{(\frac{1}{2}-\frac{\sqrt{5}}{2})}^2{x}^2-10=0$$$$x=\pm \sqrt{10+4\sqrt{5}}$$$$y=\mp \sqrt{5+\sqrt{5}}$$$$\mathit{a}=\sqrt{x^2+y^2}=\frac{\sqrt{2}}{2}(5+\sqrt{5})$$$$\mathrm{axis}\mathit{b}:y=(\frac{1}{2}+\frac{\sqrt{5}}{2})x$$$$x^2+2(\frac{1}{2}+\frac{\sqrt{5}}{2})x^2+2{(\frac{1}{2}+\frac{\sqrt{5}}{2})}^2{x}^2-10=0$$$$x=\pm \sqrt{10-4\sqrt{5}}$$$$y=\pm \sqrt{5-\sqrt{5}}$$$$\mathit{b}=\sqrt{x^2+y^2}=\frac{\sqrt{2}}{2}(5-\sqrt{5})$$$$A=\mathit{a}\mathit{b}\pi =10\pi$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com