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The values of $m$ for which the line $y=m x+2$ becomes a tangent to the hyperbola $4 x^2-9 y^2=36$ is
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Verified Answer
The correct answer is:
$\pm \frac{2 \sqrt{2}}{3}$
We have, line, $y=m x+2$... (i)
Hyperbola,
$4 x^2-9 y^2=36... (ii)$
On solving eqs. (i) and (ii), we get
$x^2\left(4-9 m^2\right)-36 m x-72=0$
Since, this line is a tangent of hyperbola
$\begin{array}{ll}
\therefore & \mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=0 \\
\therefore & (36 m)^2+4 \times 72\left(4-9 m^2\right)=0 \\
\Rightarrow & 36 \times 36 m^2+4 \times 36 \times 2\left(4-9 m^2\right)=0 \\
\Rightarrow & 9 m^2+8-18 m^2=0 \\
\Rightarrow & 9 m^2=8 \\
\Rightarrow & m^2=\frac{8}{9} \\
\therefore & m^2= \pm \sqrt{\frac{8}{9}}= \pm \frac{2 \sqrt{2}}{3}
\end{array}$
Hyperbola,
$4 x^2-9 y^2=36... (ii)$
On solving eqs. (i) and (ii), we get
$x^2\left(4-9 m^2\right)-36 m x-72=0$
Since, this line is a tangent of hyperbola
$\begin{array}{ll}
\therefore & \mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=0 \\
\therefore & (36 m)^2+4 \times 72\left(4-9 m^2\right)=0 \\
\Rightarrow & 36 \times 36 m^2+4 \times 36 \times 2\left(4-9 m^2\right)=0 \\
\Rightarrow & 9 m^2+8-18 m^2=0 \\
\Rightarrow & 9 m^2=8 \\
\Rightarrow & m^2=\frac{8}{9} \\
\therefore & m^2= \pm \sqrt{\frac{8}{9}}= \pm \frac{2 \sqrt{2}}{3}
\end{array}$
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