What is (x^2}{25}+\frac{y^2)/9 <1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2}{25}+\frac{y^2)/9 <1

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

(h, k) = (0, 0), a = 5, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 5, b = 3

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

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

y = 4 - \frac{x ^{2}}{4 y}

3 x^{2} + y^{2} - 12 = 0

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

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