Question

If the straight line $y=mx+2,m\in R$ divides the region bounded by $9x^2+4y^2-16y-20=0$ into two parts, then the ratio of the area of these parts can be

• A. $1:4$ for $m=-2$
• B. $1:1$ for $m=-1$
• C. $1:2$ for $m=3$
• D. $2:1$ for $m=2$

Answer 1

The correct option is B $1:1$ for $m=-1$

Given curve : $9x^2+4y^2-16y-20=0$
Can be written as,
$\frac{x^2}{4}+\frac{(y-2{)}^2}{9}=1$
which is an ellipse with centre $(0,2)$
and the straight line $y=mx+2$ passes through the centre of the ellipse.
So the ratio of area will be $1:1$ for all $m\in R$