What is the vertices 2((x-6)^2)/3+7((y+4)^2)/4 =1 ?

The vertices 2((x-6)^2)/3+7((y+4)^2)/4 =1 is (6+sqrt(3/2),-4),(6-sqrt(3/2),-4)

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vertices 2((x-6)^2)/3+7((y+4)^2)/4 =1

vertices

2 \frac{( x - 6 ) ^{2}}{3} + 7 \frac{( y + 4 ) ^{2}}{4} = 1

(6 + \frac{3}{2}, - 4), (6 - \frac{3}{2}, - 4)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

2 \frac{( x - 6 ) ^{2}}{3} + 7 \frac{( y + 4 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (6, - 4), a = \frac{3}{2}, b = \frac{2}{7}

(6 + \frac{3}{2}, - 4), (6 - \frac{3}{2}, - 4)

\frac{x ^{2}}{96} + \frac{y ^{2}}{100} = 1

\frac{( x + 4 ) ^{2}}{24} + \frac{( y + 2 ) ^{2}}{49} = 1

\frac{( x ^{2} )}{9} + \frac{( y ^{2} )}{4} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{9} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{49} = 1

What is the vertices 2((x-6)^2)/3+7((y+4)^2)/4 =1 ?
The vertices 2((x-6)^2)/3+7((y+4)^2)/4 =1 is (6+sqrt(3/2),-4),(6-sqrt(3/2),-4)