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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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A common tangent to $9x^2-16y^2=144$ and $x^2+y^2=9$ is

$\begin{array}{1 1}(A)\;y=\frac{3}{\sqrt 7} x +\frac{15}{\sqrt 7} \\(B)\;y= 3 \sqrt {\frac{2}{7}} x+ \frac{15}{\sqrt 7} \\(C)\;y=2 \sqrt {\frac{3}{7}} x +15 \sqrt 7 \\(D)\;\text{none of these } \end{array}$

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Let $y=mx+c$ be a common tangent to $9x^2-16y^2=144$ and $x^2+y^2=9$
Since $y=mx+c$ is a tangent to $ 9x^2-16y^2=144$
$\therefore \;c^2=a^2m^2-b^2$
$c^2=16m^2-9$ -----(1)
Now , $y=mx+c$ is a tangent to
$\therefore \large\frac{c}{\sqrt {m^2+1}}$$=3$
=> $ c^2=9(1+m^2)$-----(2)
From (1) and (2) we get
$16m^2-9=9 + 9m^2$
=> $m =\pm 3 \sqrt {\frac{2}{7}}$
Substitute the value of m in eq (2) we get,
$c= \pm \large\frac{15}{\sqrt 7}$
Hence $y= 3 \sqrt {\large\frac{2}{7} }$$x+\large\frac{15}{\sqrt {7}}$ is a common tangent
Hence B is the correct answer.
answered Apr 11, 2014 by meena.p

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