Complex Numbers and Coordinate Geometry

Algebra Level 3

The equation : ∣ z − 1 ∣ + ∣ z + 1 ∣ = 4 |{ z }-1|+|{ z }+1|=4 represents an ellipse having eccentricity :

1/3 1/4 2/3 1/2

Sahil Bansal by Sahil Bansal , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
Jul 6, 2019

Let z = x + y i z = x + yi be any complex number. Substituting this value into the above modulus equation yields:

∣ ( x − 1 ) + y i ∣ + ∣ ( x + 1 ) + y i ∣ = 4 ; |(x-1) + yi| + |(x+1) + yi| = 4;

or ( x − 1 ) 2 + y 2 + ( x + 1 ) 2 + y 2 = 4 ; \sqrt{(x-1)^2 + y^2} + \sqrt{(x+1)^2 + y^2} = 4;

or ( x − 1 ) 2 + 2 y 2 + ( x + 1 ) 2 + 2 ⋅ ( x − 1 ) 2 y 2 + ( x + 1 ) 2 y 2 + y 4 + ( x 2 − 1 ) 2 = 16 ; (x-1)^2 + 2y^2 + (x+1)^2 + 2 \cdot \sqrt{(x-1)^2 y^2 + (x+1)^2 y^2 + y^4 + (x^2 - 1)^2} = 16;

or 2 x 2 + 2 y 2 + 2 + 2 ⋅ 2 x 2 y 2 − 2 x y 2 + 2 y 2 + 2 x y 2 + y 4 + x 4 − 2 x 2 + 1 = 16 ; 2x^2 + 2y^2 + 2 + 2 \cdot \sqrt{2x^2 y^2 - 2xy^2 + 2y^2 + 2xy^2 + y^4 + x^4 - 2x^2 + 1} = 16;

or ( x 2 + y 2 ) 2 − 2 ( x 2 − y 2 ) + 1 = 7 − x 2 − y 2 ; \sqrt{(x^2 + y^2)^2 - 2(x^2 - y^2) + 1} = 7 - x^2 - y^2;

or ( x 2 + y 2 ) 2 − 2 ( x 2 − y 2 ) + 1 = 49 − 14 ( x 2 + y 2 ) + ( x 2 + y 2 ) 2 ; (x^2 + y^2)^2 - 2(x^2 - y^2) + 1 = 49 - 14(x^2 + y^2) + (x^2 + y^2)^2;

or 12 x 2 + 16 y 2 = 48 ; 12x^2 + 16y^2 = 48;

or x 2 4 + y 2 3 = 1. \frac{x^2}{4} + \frac{y^2}{3} = 1.

Taking a = 2 , b = 3 a = 2, b = \sqrt{3} the focal distance computes to c = a 2 − b 2 = 4 − 3 = 1 c = \sqrt{a^2 - b^2} = \sqrt{4-3} = 1 . The eccentricity is now computed to be e = c a = 1 2 . e = \frac{c}{a} = \boxed{\frac{1}{2}}.

0 Helpful 0 Interesting 0 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...