Question

Foci of the ellipse $25x^2+9y^2-150x-90y+225=0$ are :

• A. $(3,9),(3,1)$
• B. $(4,9),(4,1)$
• C. $(5,7),(4,7)$
• D. $(6,9),(5,1)$

Answer 1

The correct option is A $(3,9),(3,1)$

Given,

$25x^2+9y^2-150x-90y+225=0$

$25(x^2-6x)+9(y^2-10y)=-225$

$(x^2-6x)+\frac{9}{25}(y^2-10y)=-9$

$\frac{{(x-3)}^2}{9}+\frac{{(y-5)}^2}{25}=1$

$\frac{{(x-3)}^2}{3^2}+\frac{{(y-5)}^2}{5^2}=1$

$(h,k)=(3,5),a=3,b=5$

$c=\sqrt{b^2-a^2}$

Foci $(h,k+c),(h,k-c)$

$c=\sqrt{5^2-3^2}:4$

$(3,5+4),(3,5-4)$

$(3,9),(3,1)$