A) Parabola
B) Ellipse
C) Hyperbola
D) Circle
Correct Answer: B
Solution :
\[|z-2|+|z+2|\ =8\] Þ \[\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}+\sqrt{{{(x+2)}^{2}}+{{y}^{2}}}=8\] Þ \[{{x}^{2}}+{{y}^{2}}+4-4x=64+{{x}^{2}}+{{y}^{2}}+4+4x\]\[-16\sqrt{{{(x+2)}^{2}}+{{y}^{2}}}\] Þ \[-8x-64=-16\sqrt{{{(x+2)}^{2}}+{{y}^{2}}}\] Þ \[(x+8)=2\sqrt{{{(x+2)}^{2}}+{{y}^{2}}}\] Þ \[{{x}^{2}}+64+16x=4[{{x}^{2}}+{{y}^{2}}+4+4x]\] Þ \[3{{x}^{2}}+4{{y}^{2}}-48=0\] \[\Rightarrow \frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{12}=1\], which is an ellipse.
You need to login to perform this action.
You will be redirected in
3 sec
