What is (9x^2)/(25)+(16y^2)/(25)<1 ?

The solution to (9x^2)/(25)+(16y^2)/(25)<1 is Ellipse with (h,k)=(0,0),a= 5/3 ,b= 5/4

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(9x^2)/(25)+(16y^2)/(25)<1

\frac{9 x ^{2}}{25} + \frac{16 y ^{2}}{25} < 1

(h, k) = (0, 0), a = \frac{5}{3}, b = \frac{5}{4}

Solution steps

\frac{9 x ^{2}}{25} + \frac{16 y ^{2}}{25} < 1

Rewrite

\frac{9 x ^{2}}{25} + \frac{16 y ^{2}}{25} < 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{5}{3} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{5}{4} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{5}{3}, b = \frac{5}{4}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = \frac{5}{3}, b = \frac{5}{4}

2 x^{2} + y^{2} = 3

49 x^{2} + 16 y^{2} = 784

\frac{( x - 6 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

x^{2} + 4 y^{2} - 2 x - 16 y + 13 = 0

9 x^{2} + 4 y^{2} - 18 x + 8 y + 4 = 0

What is (9x^2)/(25)+(16y^2)/(25)<1 ?
The solution to (9x^2)/(25)+(16y^2)/(25)<1 is Ellipse with (h,k)=(0,0),a= 5/3 ,b= 5/4