Browse Questions

# Find the position of the point (4,-3) relative to the ellipse $5x^2+7y^2=140$?

$\begin{array}{1 1}(a)\;\text{Centre of ellipse}\\(b)\;\text{Inside the ellipse}\\(c)\;\text{Outside the ellipse}\\(d)\;\text{None of these}\end{array}$

$5(4)^2+7(-3)^2-140=3 > 0$
So the point (4,-3) lies outside the ellipse $5x^2+7y^2=140$
Hence (c) is the correct answer.