differentiation formula

Question

for x^2 /169 + y^2 /144+1
find
a. length of semi major axis
b. " " " minor axis
c. coordinates of the foci
d. eccentricity

Answer 1

Let  us assume that x ² / 169 + y ² / 144    = 1

Identify the type of conic and orientation.

There is both an x ² and a  y ², so it is an ellipse or hyperbola.

The x ² and y ² have the same sign, so it is an ellipse.

Compare the size of the denominators: The denominator of x ² is larger, so the ellipse is horizontal.

General equation of a horizontal ellipse:

(x - h) ² / a ² + (y - k) ² / b ² = 1

x ² / 169 + y ² / 144    = 1

compare to the general equation of a horizontal ellipse.

h = 0 , k = 0

center (0, 0)

a ² = 169 ,  b ² = 144

linear eccentricity c = √(a ² - b ²)

foci ( h ± c, k )

a)

a ² = 169

length of semi-major axis = a = 13

b)

b ² = 144

length of semi-minor axis = b = 12

c)

linear eccentricity c = √( a ² -  b ²) =√( 169 -144) = √( 25) =  5.

d)

foci ( 0 ± c , 0) = (- 5 , 0) and ( 5 , 0 ).

Comments

check your answer,

TheFormula for the Eccentricy of an Ellipse = c / a, where c is the distance from the center to the focus of the ellipse a is the distance from the center to a vertex.

The relation between a, b and c is b² = c² - a².

Therefore c = √( a ² -  b ²) =√( 169 -144) = √( 25) =  5.

The Eccentricy of an Ellipse is c / a = 5/13.