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The equation of the tangent of the hyperbola 4 x^2-9 y^2=1 which is parallel to the line 4 y=5 x+27 is 24 y-30 x=k, where k is equal to

Option: 1

\pm \sqrt{161}


Option: 2

\pm \sqrt{160}
 


Option: 3

\pm \sqrt{170}
 


Option: 4

None of these 


Answers (1)

best_answer

(a) The given equation can be written as 

             \frac{x^2}{1 / 4}-\frac{y^2}{1 / 9}=1

So that           a^2=\frac{1}{4} \text { and } b^2=\frac{1}{9}

We know that,  y=m x+\sqrt{a^2 m^2-b^2} is always a tangent to the hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 for all m.

In our case y=m x+\sqrt{\frac{m^2}{4}-\frac{1}{9}} is a tangent to the given hyperbola.

But given equation of tangent is

\begin{aligned} 24 y & =30 x+k \\ \Rightarrow \quad y & =\frac{30}{24} x+\frac{k}{24}=\frac{5}{4} x+\frac{k}{24} \end{aligned}

is a tangent to the given hyperbola.
Hence, identifying the two equations, we get

\begin{aligned} & m=\frac{5}{4} \text { and } \frac{m^2}{4}-\frac{1}{9}=\frac{k^2}{24^2} \\ \\\Rightarrow \quad \frac{25}{64}-\frac{1}{9} & =\frac{k^2}{576} \\ \\\Rightarrow \quad \frac{161}{64 \times 9} & =\frac{k^2}{576} \end{aligned}

\begin{aligned} \\\Rightarrow \quad k^2 & =\frac{161 \times 576}{64 \times 9}=161 \\ \\k & = \pm \sqrt{161} \end{aligned}

Posted by

Shailly goel

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