question_answer

Consider any point P on the ellipse \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\] in the first quadrant. Let r and s represent its distances from (4, 0) and (-4, 0) respectively, then (r + s) is equal to

A) 10 unit

B) 9 unit

C) 8 unit

D) 6 unit

Correct Answer: A

Solution :

[a] \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\]

Put \[x=3\]

\[\frac{9}{25}+\frac{{{y}^{2}}}{9}=1\]

\[y=\frac{12}{5}\]

\[P=(3,12/5)\]

\[r=PO=\sqrt{{{(4-3)}^{2}}+{{\left( 0-\frac{12}{5} \right)}^{2}}}\]

\[=17/5\]

\[S=PO'=\sqrt{[-4-3]+{{\left( 0-\frac{12}{5} \right)}^{2}}}\]

\[=33/5\]

\[r+s=\frac{17}{5}+\frac{33}{5}=\frac{50}{5}=10\,\,units\]