Question

The abscissae of the points on the ellipse $9x^2+25y^2-18x-100y-116=0$ lie between:

• A. $3,-5$
• B. $-4,6$
• C. $-5,7$
• D. $2,5$

Answer 1

Answer: (B)

$-4,6$

We are given an ellipse $9x^2+25y^2-18x-100y-116=0$

This can be written as $9x^2-18x+25y^2-100y-116=0\Rightarrow 9(x^2-2x)+25(y^2-4y)=116$

Creating perfect squares we get,

$\Rightarrow 9{(x-1)}^2-9+25{(y-2)}^2-100=116\Rightarrow 9{(x-1)}^2+25{(y-2)}^2=225$

This can be written as $\frac{{(x-1)}^2}{25}+\frac{{(y-2)}^2}{9}=1$

This is centered at $O:(1,2)$ and $a=5,b=3$

So extremes of abscissae will be $1-5$ to $1+5$ or $-4,6$