What is (x-7/4)+(y+3/4)^2= 169/16 ?

The solution to (x-7/4)+(y+3/4)^2= 169/16 is Parabola with (h,k)=(197/16 ,-3/4),p=-1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-7/4)+(y+3/4)^2= 169/16

(x - \frac{7}{4}) + (y + \frac{3}{4})^{2} = \frac{169}{16}

(h, k) = (\frac{197}{16}, - \frac{3}{4}), p = - \frac{1}{4}

Solution steps

x - \frac{7}{4} + (y + \frac{3}{4})^{2} = \frac{169}{16}

Rewrite

x - \frac{7}{4} + (y + \frac{3}{4})^{2} = \frac{169}{16}

in the standard form:

4 (- \frac{1}{4}) (x - \frac{197}{16}) = (y - (- \frac{3}{4}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{197}{16}, - \frac{3}{4}), p = - \frac{1}{4}

y^{2} - 10 x + y = 0

\frac{a - 1}{b} = \frac{a + 1}{b} + \frac{b - 1}{a}

x = (y + 3)^{2} - 7

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

y - x^{2} + 5 = 0

What is (x-7/4)+(y+3/4)^2= 169/16 ?
The solution to (x-7/4)+(y+3/4)^2= 169/16 is Parabola with (h,k)=(197/16 ,-3/4),p=-1/4