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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the condition of tangency for a line $y=mx+c$ touches the ellipse

$\begin{array}{1 1}(a)\;y=mx\pm \sqrt{a^2m^2+b^2}\\(b)\;y=mx\pm \sqrt{a^2m^2-b^2}\\(c)\;y=mx\pm \sqrt{am^2+b^2}\\(d)\;\text{None of these}\end{array}$

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$\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
$\large\frac{x^2}{a^2}+\frac{(mx+c)^2}{b^2}$$=1$
$(a^2m^2+b^2)x^2+2mca^2x+c^2a^2-a^2b^2=0$
Now if line $y=mx+c$ is tangent than discriminant of the quadratic equation should be 0
Hence $(2mca^2)^2=4\times (c^2a^2-a^2b^2)(a^2m^2+b^2)$
$4m^2c^2a^4=4(c^2a^4m^2+c^2a^2b^2-a^4b^2m^2-a^2b^4)$
$c^2a^2b^2=a^4b^2m^2+a^2b^4$
$c^2=a^2m^2+b^2$
$c=\pm \sqrt{a^2m^2+b^2}$
So line $y=mx+c$ touches the ellipse
$\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ if $c^2=a^2m^2+b^2$
Substituting c we get,
$y=mx\pm \sqrt{a^2m^2+b^2}$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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