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

Find the position of the point (1,1) relative to the ellipse $\large\frac{x^2}{64}+\frac{y^2}{100}$=1.

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

1 Answer

Given equation lof the ellipse can be written as
$100x^2+64y^2=6400$
$\Rightarrow\:100x^2+64y^2-6400=0$
Substituting the coordinates $(1,1)$ in this equation, we get
$100(1)+64(1)-6400 < 0$
Which lies outside the ellipse.
Hence (a) is the correct answer.
answered Feb 6, 2014 by sreemathi.v
edited Mar 24, 2014 by rvidyagovindarajan_1
 
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