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Question Number 27870 by ajfour last updated on 15/Jan/18

Commented by ajfour last updated on 15/Jan/18

$$Q.27818anothersolution$$

Answered by ajfour last updated on 15/Jan/18

$$translatingtheparabolay={Ax}^2$$$$to(h,k)\equiv (a+a\cos 2\theta ,a\sin 2\theta )$$$$where\tan \theta =\frac{a}{b}$$$$\Rightarrow (h,k)\equiv [a+\frac{a(b^2-a^2)}{b^2+a^2},\frac{2a^{2}b}{b^2+a^2}]$$$$\equiv [\frac{2{ab}^2}{a^2+b^2},\frac{2a^{2}b}{a^2+b^2}]$$$$y-k=A{(x-h)}^2$$$$letuscallthis$$$$v=k+A{(u-h)}^{2}sayv=f\left(u\right)$$$$rotatingitthenby2\theta about(h,k)$$$$k+(y-k)\cos 2\theta -(x-h)\sin 2\theta$$$$=k+A{[h+(x-h)\cos 2\theta +(y-k)\sin 2\theta -h]}^2$$$$\Rightarrow (y-\frac{2a^{2}b}{a^2+b^2})\left(\frac{b^2-a^2}{a^2+b^2}\right)-(x-\frac{2{ab}^2}{a^2+b^2})\left(\frac{2ab}{a^2+b^2}\right)$$$$=A{[(x-\frac{2{ab}^2}{a^2+b^2})\left(\frac{b^2-a^2}{a^2+b^2}\right)+(y-\frac{2a^{2}b}{a^2+b^2})\left(\frac{2ab}{a^2+b^2}\right)]}^2$$

Commented by mrW2 last updated on 15/Jan/18

$$Veryfinesir!$$

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