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The locus of foot of perpendicular from the focus to a tangent of the curve $ 16x^{2}+25y^{2}=400$ is

\[\begin{array}{1 1}(1)x^{2}+y^{2}=4&(2)x^{2}+y^{2}=25\\(3)x^{2}+y^{2}=16&(4)x^{2}+y^{2}=9\end{array}\]

1 Answer

$16x^2+25y^2=400$
$\large\frac{16x^2}{400}+\frac{25y^2}{400}$$=1$
$\large\frac{x^2}{25}+\frac{y^2}{16}$$=1$
$a^2= 25$
$b^2=16$
$x^2+y^2=a^2$
$\therefore $ The required locus is $x^2+y^2=25$
Hence 2 is the correct answer.
answered May 16, 2014 by meena.p
 

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