What is (x^2}{25}-\frac{(y-3)^2)/4 =1 ?

The solution to (x^2}{25}-\frac{(y-3)^2)/4 =1 is Hyperbola with (h,k)=(0,3),a=5,b=2

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-3)^2)/4 =1

\frac{x ^{2}}{25} - \frac{( y - 3 ) ^{2}}{4} = 1

(h, k) = (0, 3), a = 5, b = 2

Solution steps

\frac{x ^{2}}{25} - \frac{( y - 3 ) ^{2}}{4} = 1

Rewrite

\frac{x ^{2}}{25} - \frac{( y - 3 ) ^{2}}{4} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (0, 3), a = 5, b = 2

y^{2} - 324 x^{2} = - 36

\frac{( y ^{2} )}{27} - \frac{( x ^{2} )}{0} = 1

(x - 3)^{2} - \frac{y ^{2}}{9} = 1

8 x^{2} - 9 y^{2} - 63 x - 18 y + 65 = 0

s^{2} = d^{2} + 2^{2}

What is (x^2}{25}-\frac{(y-3)^2)/4 =1 ?
The solution to (x^2}{25}-\frac{(y-3)^2)/4 =1 is Hyperbola with (h,k)=(0,3),a=5,b=2