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If $L_1=0$ and $L_2=0$ are the asymptotes of the hyperbola $9 x^2-4 y^2+36 x+8 y-4=0$, then the product of the perpendicular distances from the point $(1,1)$ to the lines, $L_1=0$ and $L_2=0$ is
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Verified Answer
The correct answer is:
$\frac{81}{13}$
Equation of given hyperbola is
$\begin{aligned} 9 x^2-4 y^2+36 x+8 y-4 & =0 \\ \Rightarrow \quad 9\left(x^2+4 x+4\right)-4\left(y^2-2 y+1\right) & =36\end{aligned}$
$\Rightarrow \quad 9(x+2)^2-4(y-1)^2=36$
Now, combined equation of the asymptotes of the hyperbola (i) is
$\begin{array}{rlrl}9(x+2)^2 & =4(y-1)^2 \\ \Rightarrow & & 2(y-1) & = \pm 3(x+2) \\ \Rightarrow & 3 x-2 y+8 & =0 \text { and } 3 x+2 y+4=0\end{array}$
Now, product of the perpendicular distances from the point $(1,1)$ to the asymptotes is
$\frac{3-2+8}{\sqrt{13}} \times \frac{3+2+4}{\sqrt{13}}=\frac{9 \times 9}{13}=\frac{81}{13}$
Hence, option (c) is correct.
$\begin{aligned} 9 x^2-4 y^2+36 x+8 y-4 & =0 \\ \Rightarrow \quad 9\left(x^2+4 x+4\right)-4\left(y^2-2 y+1\right) & =36\end{aligned}$
$\Rightarrow \quad 9(x+2)^2-4(y-1)^2=36$
Now, combined equation of the asymptotes of the hyperbola (i) is
$\begin{array}{rlrl}9(x+2)^2 & =4(y-1)^2 \\ \Rightarrow & & 2(y-1) & = \pm 3(x+2) \\ \Rightarrow & 3 x-2 y+8 & =0 \text { and } 3 x+2 y+4=0\end{array}$
Now, product of the perpendicular distances from the point $(1,1)$ to the asymptotes is
$\frac{3-2+8}{\sqrt{13}} \times \frac{3+2+4}{\sqrt{13}}=\frac{9 \times 9}{13}=\frac{81}{13}$
Hence, option (c) is correct.
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