Explanation:
B Given,
\(16(x-2)^2+25(y+3)^2=400\)
\(\frac{(x-2)^2}{25}+\frac{(y+3)^2}{16}=1\)
Where, \(\mathrm{X}=\mathrm{x}-2\) and \(\mathrm{Y}=\mathrm{y}+3\)
Here, \((\mathrm{a}>\mathrm{b})\) then -
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{16}{25}}\)
\(\mathrm{e}=\frac{3}{5}\)
\(\text { Foci }( \pm \text { ae }, 0)=( \pm 3,0)\)
\(\therefore \mathrm{x}-2= \pm 3 \text { and } \mathrm{y}+3=0\)
\(\mathrm{x}=(5,-1) \text { and } \mathrm{y}=-3\)
Foci are \((-1,-3)\) and \((5,-3)\)
Now, distance between \((2,-3)\) and \((-1,-3)\) is
\(=\sqrt{(2+1)^2+(-3+3)^2}=3\)
And distance between \((2,-3)\) and \((5,-3)\)
\(\sqrt{(2-5)^2+(-3+3)^2}=3\)
Therefore, the sum of the distance from the foci of a point \((2,-3)\) is -
\(=3+3=6 \text { unit }\)