The distance between the tangent lines to the hyperbola $x^2-2 y^2=18$ which are perpendicular to the line…
The distance between the tangent lines to the hyperbola $x^2-2 y^2=18$ which are perpendicular to the line $y=x$ is
- 6
- $2 \sqrt{3}$
- $3 \sqrt{2}$
- 0
Solution
The line perpendicular to line $y=x$ is
$
y=-x+c \text { or } x+y-c=0
$
Given, hyperbola is $x^2-2 y^2=18$
or $\quad \frac{x^2}{18}-\frac{y^2}{9}=1$
Here, $a^2=18, b^2=9$
Condition for tangency of hyperbola is
$
\begin{aligned}
& c^2=a^2 m^2-b^2 \\
& c^2=18 \times(-1)^2-9 \\
& c^2=9 \Rightarrow c=+3
\end{aligned}
$
Equation of tangents are
$
x+y \pm 3=0
$
$
\begin{aligned}
\text { Distance between tangent } & =\frac{|6|}{\sqrt{1^2+1^2}} \\
& =\frac{6}{\sqrt{2}} \text { or } 3 \sqrt{2} \text { units }
\end{aligned}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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