Question

Foci of the ellipse $25x^2+9y^2+50x+36y-164=0$ are

• A. $(-1,2),(-1,-6)$
• B. $(-2,1),(-2,6)$
• C. $(3,-2),(-5,-2)$
• D. $(-3,2)(5,2)$

Answer 1

The correct option is A $(-1,2),(-1,-6)$

Equation of ellipse : ${25x}^2+{9y}^2+50x+36y-164=0$

${25x}^2+50x+{9y}^2+36y=164$

$25(x^2+2x)+25-25+9(y^2+4y)+36-36=164$

$25{(x+1)}^2-25+9{(y+2)}^2-36=164$

$25{(x+1)}^2+9{(y+2)}^2=164+25+36$

$25{(x+1)}^2+9{(y+2)}^2=225$

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

$h=1,k=-2,a=3,b=5$

Centre of ellipse, $(h,k)=(1-,-2)$

Distance from centre to focus = $\sqrt{a^2-b^2}=\sqrt{25-9}=4$

vertex $=(h,k+a)$ and $(h,k-a)$

=$(-1,-2+3)$and$(-2,-2-3)$

Focii $=(h,k+c)$ and $(h,k-c)$

= $(-1,-1+4)$ and$(-1,-1-4)$

= $(-1,3)$and$(-1,-5)$ Ans.