logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

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

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

Can you answer this question?
 
 

1 Answer

0 votes
$5(4)^2+7(-3)^2-140=3 > 0$
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
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...