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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

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

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

1 Answer

$5(4)^2+7(-3)^2-140=3 > 0$
So the point (4,-3) lies outside the ellipse $5x^2+7y^2=140$
Hence (c) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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