• # question_answer The minimum and maximum distances of a point (3, 12) from the ellipse$25{{x}^{2}}+9{{y}^{2}}-150x-90y+225=0$ are $a$ and b,  then a and b satisfy the relation. A)  $a+b=10$                B)  $|b-a|\,=10$ C)  $|a\pm b|\,=2009$  D)  $|b-a|\,=4$

The give equation of the ellipse can be written as $25\,{{(x-3)}^{2}}+9\,{{(y-5)}^{2}}\,=225$ $\Rightarrow$            $\frac{{{(x-3)}^{2}}}{9}+\frac{{{(y-5)}^{2}}}{25}=1$ Which shows that the centre of ellipse is (3, 5) and vertices of the ellipse are (3, 0) and (3, 10). As (3, 10) is vertex of the ellipse and point (3, 12) lies outside the ellipse. $\therefore$    $a=$ Minimum distance = Distance between (3, 10) and (3, 12) = 2 and b = Maximum distance = Distance between (3, 0) and (3, 12) = 12 $\therefore$    $a+b=14$     and      $|b-a|=10$