## Solving Lesson 82 Page 99 SBT Math 10 – Kite>

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 |MFfirstMF2| = 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 |MFfirstMF2| = 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 |MFfirstMF2| = 4 is the hyperbolic curve (H)

We have: |MFfirstMF2| = 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: $$\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{12}} = 1$$