JEE Main On 8 April 2017 Question 7
Question: Let $ z\in C, $ the set of complex numbers. Then the equation, $ 2|z+3i|-|z-i|=0 $ a circle with radius $ \frac{8}{3}. $ [JEE Online 08-04-2017]
Options:
A) a circle with radius $ \frac{8}{3} $
B) an ellipse with length of minor axis $ \frac{16}{9} $
C) an ellipse with length of major axis $ \frac{16}{3} $
D) a circle with diameter $ \frac{10}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 2|x+i(y+3)|=|x+i(y-1)| $ $ =2\sqrt{x^{2}+{{(y+3)}^{2}}}\sqrt{x^{2}+{{(y-1)}^{2}}} $ $ =4x^{2}+4{{(y+3)}^{2}}=x^{2}+{{(y-1)}^{2}} $ $ =3x^{2}=y^{2}-2y+1-4y^{2}-24y-36 $ $ =3x^{2}+3y^{2}+26y+35=0 $ $ =x^{2}+y^{2}+\frac{26}{3}y+\frac{35}{3}=0 $ $ =r=\sqrt{0+\frac{169}{9}-\frac{35}{3}} $ $ =\sqrt{\frac{64}{9}}=\frac{8}{3} $
