Question Number 214675 by cherokeesay last updated on 16/Dec/24
Answered by mr W last updated on 16/Dec/24
Commented by mr W last updated on 16/Dec/24
![x^2 +y^2 =2^2 =4 AB=(√(x^2 +9y^2 ))=(√(x^2 +9(4−x^2 )))=(√(36−8x^2 )) CD=(√(y^2 +9x^2 ))=(√(4−x^2 +9x^2 ))=(√(4+8x^2 )) (p/(AB))=(q/(CD))=(2/(6−2))=(1/2) p=((√(36−8x^2 ))/2)=(√(9−2x^2 )) q=((√(4+8x^2 ))/2)=(√(1+2x^2 )) p^2 +q^2 −2pq cos 45°=2^2 9−2x^2 +1+2x^2 −(√(2(9−2x^2 )(1+2x^2 )))=4 (√(2(9−2x^2 )(1+2x^2 )))=6 4x^4 −16x^2 +9=0 x^2 =2+((√7)/2) y^2 =2−((√7)/2) xy=(√((2+((√7)/2))(2−((√7)/2))))=(3/2) [ABCD]=((4x×4y)/2)=8xy=8×(3/2)=12](https://www.tinkutara.com/question/Q214686.png)
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${AB}=\sqrt{{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} +\mathrm{9}\left(\mathrm{4}−{x}^{\mathrm{2}} \right)}=\sqrt{\mathrm{36}−\mathrm{8}{x}^{\mathrm{2}} } \\ $$$${CD}=\sqrt{{y}^{\mathrm{2}} +\mathrm{9}{x}^{\mathrm{2}} }=\sqrt{\mathrm{4}−{x}^{\mathrm{2}} +\mathrm{9}{x}^{\mathrm{2}} }=\sqrt{\mathrm{4}+\mathrm{8}{x}^{\mathrm{2}} } \\ $$$$\frac{{p}}{{AB}}=\frac{{q}}{{CD}}=\frac{\mathrm{2}}{\mathrm{6}−\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}=\frac{\sqrt{\mathrm{36}−\mathrm{8}{x}^{\mathrm{2}} }}{\mathrm{2}}=\sqrt{\mathrm{9}−\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${q}=\frac{\sqrt{\mathrm{4}+\mathrm{8}{x}^{\mathrm{2}} }}{\mathrm{2}}=\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{pq}\:\mathrm{cos}\:\mathrm{45}°=\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{9}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} −\sqrt{\mathrm{2}\left(\mathrm{9}−\mathrm{2}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)}=\mathrm{4} \\ $$$$\sqrt{\mathrm{2}\left(\mathrm{9}−\mathrm{2}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)}=\mathrm{6} \\ $$$$\mathrm{4}{x}^{\mathrm{4}} −\mathrm{16}{x}^{\mathrm{2}} +\mathrm{9}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}+\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$${xy}=\sqrt{\left(\mathrm{2}+\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)\left(\mathrm{2}−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left[{ABCD}\right]=\frac{\mathrm{4}{x}×\mathrm{4}{y}}{\mathrm{2}}=\mathrm{8}{xy}=\mathrm{8}×\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{12}\: \\ $$
Commented by cherokeesay last updated on 16/Dec/24
$${great}\:!\:{thank}\:{you}\:{master}\:! \\ $$
Commented by som(math1967) last updated on 17/Dec/24
$${Sir}\:{angle}\:{between}\:\mathrm{3}{x}\:{and}\:\mathrm{3}{y}\:{is} \\ $$$$\mathrm{90}°\:{so}\:{meeting}\:{point}\:{of}\:{them}\:{is} \\ $$$${circumcentre}\:{of}\:{triangle}\: \\ $$$${so}\:\mathrm{3}{x}=\mathrm{3}{y}\:? \\ $$
Commented by mr W last updated on 17/Dec/24
$${no}\:{sir}.\:{it}\:{means}\:{only}\:{that}\:{the}\:{meeting} \\ $$$${point}\:{is}\:{on}\:{the}\:{circumcircle}\:{with} \\ $$$${AD}\:{as}\:{diameter}. \\ $$
Commented by som(math1967) last updated on 17/Dec/24
$${Ok}\:{sir}\:{thank}\:{you} \\ $$