What is the foci-2/3 y^2=(x-4)(x-1) ?

The foci-2/3 y^2=(x-4)(x-1) is (5/2 , 3/(2sqrt(2))),(5/2 ,-3/(2sqrt(2)))

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foci-2/3 y^2=(x-4)(x-1)

foci

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

(\frac{5}{2}, \frac{3}{2 2}), (\frac{5}{2}, - \frac{3}{2 2})

Solution steps

(h, k + c), (h, k - c)

Calculate ellipse properties

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

Ellipse with center

(h, k) = (\frac{5}{2}, 0), b = \frac{3 3}{2 2}, a = \frac{3}{2}

(\frac{5}{2}, 0 + c), (\frac{5}{2}, 0 - c)

Compute

c :

c = (\frac{3 3}{2 2})^{2} - (\frac{3}{2})^{2} : \frac{3}{2 2}

(\frac{5}{2}, 0 + \frac{3}{2 2}), (\frac{5}{2}, 0 - \frac{3}{2 2})

Refine

(\frac{5}{2}, \frac{3}{2 2}), (\frac{5}{2}, - \frac{3}{2 2})

\frac{x ^{2}}{5} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{8} = 1

9 x^{2} + 4 y^{2} - 54 x + 40 y + 145 = 0

9 x^{2} + y^{2} + 36 x - 2 y + 1 = 0

\frac{( x ) ^{2}}{46} + \frac{( y + 8 ) ^{2}}{26} = 1

What is the foci-2/3 y^2=(x-4)(x-1) ?
The foci-2/3 y^2=(x-4)(x-1) is (5/2 , 3/(2sqrt(2))),(5/2 ,-3/(2sqrt(2)))