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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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For what value of $\lambda$ does the line $y=x+\lambda$ touches the ellipse $\large\frac{x^2}{36}+\frac{y^2}{25}$$=1$?

$\begin{array}{1 1}(a)\;\sqrt{62}\\(b)\;\sqrt{61}\\(c)\;\sqrt{71}\\(d)\;\sqrt{82}\end{array}$

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For this line $y=x+\lambda$ to touch the ellipse $c^2=a^2m^2+b^2$
$\lambda^2=36\times 1+25$
$\lambda ^2=61$
$\lambda =\sqrt{61}$
Hence (b) is the correct answer.
answered Feb 7, 2014 by sreemathi.v

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