Explanation
$$25{ x }^{ 2 }+16{ y }^{ 2 }-150x=175\\ { (5x) }^{ 2 }-150x+{ (15) }^{ 2 }+16{ y }^{ 2 }=175+{ (15) }^{ 2 }\\ { (5x-15) }^{ 2 }+16{ y }^{ 2 }=400\\ 25{ (x-3) }^{ 2 }+16{ y }^{ 2 }=400\\ \dfrac { { (x-3) }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1$$
Major axis of the ellipse is $$y$$ axis
So $$a=5,b=4$$
$${ e }^{ 2 }=1-\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } =1-\dfrac { 16 }{ 25 } =\dfrac { 9 }{ 25 } \\ \Rightarrow e=\dfrac { 3 }{ 5 } $$
Coordinates of foci of standard ellipse with major axis as $$y$$ are $$(0,\pm ae)$$
$$\Rightarrow $$ foci of given ellipse is
$$\left( 3,\pm 5\times \dfrac { 3 }{ 5 } \right) \\ \left( 3,\pm 3 \right) $$'
So opttion $$C$$ is correct
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