**Topic**

In the coordinate plane O*xy*for two points *F*_{first}(−4 ; 0) and *F*_{2}(4 ; 0).

a) Find the equation of a circle with diameter *F*_{first}*F*_{2}

b) Set of points *USA* in the coordinate plane satisfying *MF*_{first} + *MF*_{2} = 12 is a conic (*E*). Said (*E*) is which conic and write the canonical equation of (*E*)

c) Set of points *USA* in the coordinate plane satisfying |*MF*_{first} – *MF*_{2}| = 4 is a conic (*H*). Said (*H*) is which conic and write the canonical equation of (*H*)

**Solution method – See details**

Step 1: Find the coordinates of the center and radius of the circle whose diameter is *F*_{first}*F*_{2 } then write the circle PT

Step 2: Write the canonical PT of the ellipse with 2 foci *F*_{first}(−4 ; 0), *F*_{2}(4 ; 0) and *MF*_{first} + *MF*_{2} = 12

Step 3: Write canonical PT of hyperbola with 2 foci *F*_{first}(−4 ; 0), *F*_{2}(4 ; 0) and |*MF*_{first} – *MF*_{2}| = 4

**Detailed explanation**

a) Call *I* is the midpoint of *F*_{first}*F*_{2 }\( \Rightarrow I(0;0)\)\( \Rightarrow I{F_1} = I{F_2} = 4\)

Diameter circle *F*_{first}*F*_{2} have heart *I*(0 ; 0) and radius *CHEAP* = 4 has PT: \({x^2} + {y^2} = 16\)

b) Set of points *USA* in the coordinate plane satisfying *MF*_{first} + *MF*_{2} = 12 is the ellipse (*E*)

We have: *MF*_{first} + *MF*_{2} = 12 = 2*a* \( \Rightarrow a = 6\)

\({F_1}{F_2} = 8 = 2c \Rightarrow c = 4\)

Then \({b^2} = {a^2} – {c^2} = 36 – 16 = 20\)

So the ellipse (*E*) has PT: \(\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{20}} = 1\)

b) Set of points *USA* in the coordinate plane satisfying |*MF*_{first} – *MF*_{2}| = 4 is the hyperbolic curve (*H*)

We have: |*MF*_{first} – *MF*_{2}| = 4 = 2*a* \( \Rightarrow a = 2\)

\({F_1}{F_2} = 8 = 2c \Rightarrow c = 4\)

Then \({b^2} = {c^2} – {a^2} = 16 – 4 = 12\)

So the hyperbola (*H*) has PT: \(\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{12}} = 1\)