Topic
In the coordinate plane Oxyfor two points Ffirst(−4 ; 0) and F2(4 ; 0).
a) Find the equation of a circle with diameter FfirstF2
b) Set of points USA in the coordinate plane satisfying MFfirst + MF2 = 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 |MFfirst – MF2| = 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 FfirstF2 then write the circle PT
Step 2: Write the canonical PT of the ellipse with 2 foci Ffirst(−4 ; 0), F2(4 ; 0) and MFfirst + MF2 = 12
Step 3: Write canonical PT of hyperbola with 2 foci Ffirst(−4 ; 0), F2(4 ; 0) and |MFfirst – MF2| = 4
Detailed explanation
a) Call I is the midpoint of FfirstF2 \( \Rightarrow I(0;0)\)\( \Rightarrow I{F_1} = I{F_2} = 4\)
Diameter circle FfirstF2 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 MFfirst + MF2 = 12 is the ellipse (E)
We have: MFfirst + MF2 = 12 = 2a \( \Rightarrow a = 6\)
\({F_1}{F_2} = 8 = 2c \Rightarrow c = 4\)
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 |MFfirst – MF2| = 4 is the hyperbolic curve (H)
We have: |MFfirst – MF2| = 4 = 2a \( \Rightarrow a = 2\)
\({F_1}{F_2} = 8 = 2c \Rightarrow c = 4\)
Then \({b^2} = {c^2} – {a^2} = 16 – 4 = 12\)
So the hyperbola (H) has PT: \(\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{12}} = 1\)