differentiation formula 2

Question

write the eq. of the ellipse w/ center at the origin, its foci on the x axis, which passes through the points (-3, 2 sq. root of 3) and (4, 4 sq. root of 5/ 3)

3. given the ellipse whose eq. is 9x^2+ 16y^2- 36x+96y+36=0
FIND
A. coordinates of its center
b. semi major axis
c. semi minor axis
d. foci
e. length of the latus rectum
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Answer 1

The standard ellipse formula is:

x²/a²+y²/b²=1. This is a centralised ellipse with x as the major axis and a as the length of its semi-major axis (radius), b as the length of the semi-minor axis. (-3,2√3) is a point on the ellipse.

9/a²+12/b²=1. Multiply through by 16 (we'll see why later):

144/a²+192/b²=16.

(4,4√5/3) is another point:

16/a²+80/9b²=1. Multiply through by 9:

144/a²+80/b²=9.

Now subtract one equation from the other:

112/b²=7, b²=112/7=16 so b=4.

9/a²+12/16=1, 9/a²=4/16=¼, a²=36, a=6.

x²/36+y²/4=1. The focal length c=√(a²-b²)=√(36-16)=√20=2√5.

The foci on the x-axis are at (-2√5,0) and (2√5,0).

(3)

Rewrite: 9(x²-4x+4-4)+16(y²+6y+9-9)=-36,

9(x-2)²-36+16(y+3)²-144=-36, (a) centre is (2,-3).

9(x-2)²+16(y+3)²=144,

(x-2)²/16+(y+3)²/9=1, (b) semi-major axis length a=√16=4; (c) semi-minor axis length b=√9=3.

Now, at this point we can replace the x-y plane with the X-Y plane where X=x-2 and Y=y+3 and:

X²/16+Y²/9=1 which is the centralised form of the ellipse. The reason for shifting the frame of reference is to make it easier to work out other properties of the ellipse and reduce the chances of making calculation errors.

Focal length c=√(a²-b²)=√(16-9)=√7. The foci are √7 away from the centre horizontally, so their coordinates in the X-Y plane are (-√7,0) and (√7,0) because the foci lie on the major axis which is now the X-axis, since, in the X-Y plane the centre of the ellipse is at the origin (0,0). When we use the x-y plane as the reference frame, we simply replace X with x-2 and Y with y+3. Therefore the y-coord for both foci is zero (the X-axis), Y=0=y+3, so y=-3. The X coords are X=-√7 and X=√7, so x-2=-√7, x=2-√7; and x-2=√7, x=2+√7, making (d) the foci (2-√7,-3) and (2+√7,-3).

Changing the frame of reference from x-y to X-Y doesn't change the shape or size of the ellipse, so the latus recta (LR) are still the same length whichever frame we use. Using the X-Y plane makes it easier to calculate the length of the LR (each LR has the same length). So we choose one of the foci. Let's choose X=√7 as the X coord. Plug X=√7 into X²/16+Y²/9=1, then:

7/16+Y²/9=1, Y²/9=1-7/16=9/16, Y²=81/16, Y=±9/4.

(e) The length of the LR is 9/4-(-9/4)=9/2.