Question

The distance between one focus to one end of minor axis of the ellipse $16x^2+25y^2-50y-375=0$ is:

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

Answer: (B)

$5$

We have, $16x^2+25y^2-50y-375=0$
$\Rightarrow 16x^2+25(y^2-2y)=375$
$\Rightarrow 16x^2+25(y^2-2y+1)=375+25=400$
Divide both sides by $400$ we get,
$\frac{x^2}{25}+\frac{(y-1{)}^2}{16}=1$
Comparing with standard ellipse, $a=5,b=4$
Thus, eccentricity $,e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Hence one focus is $S(ae,0)=(3,0)$ (you can take other one as well)
And one end of minor axis is $B(0,4)$ (Here also you have 2 choices )
Hence required distance $,SB=\sqrt{(3-0{)}^2+(0-4{)}^2}=5$