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  • Two tangents to the hyperbola <img alt="\mathrm{\frac{x^2}{25}-\frac{y^2}{9}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B25%7D-%5Cfrac%7By%5E2%7D%7B9%7

Two tangents to the hyperbola \mathrm{\frac{x^2}{25}-\frac{y^2}{9}=1}, having slopes 2 and m where \mathrm{(m \neq 2)} cuts the axes at four concyclic points, then the slope m is

Option: 1

-\frac{1}{4}


Option: 2

-2


Option: 3

\frac{1}{2}


Option: 4

3


Answers (1)

best_answer

Tangents having slope ' 2 ' are

\mathrm{y=2 x \pm \sqrt{100-9}=2 x \pm \sqrt{91}}

It meets the axis at \mathrm{A\left(\mp \frac{\sqrt{91}}{2}, 0\right), B(0, \pm \sqrt{91}).}

Tangents having slope ' m ' are \mathrm{y=m x \pm \sqrt{25 m^2-9}}

It meets the axis at \mathrm{D\left( \pm \frac{\sqrt{25 m^2-9}}{m}, 0\right), C\left(0, \mp \sqrt{25 m^2-9}\right)}

Since, ABCD is a cyclic quadrilateral.

\mathrm{ \therefore \quad O A \times O D=O B \times O C }

                                                     [where O is the centre of hyperbola]

\mathrm{ \begin{aligned} & \Rightarrow \frac{\sqrt{91} \sqrt{25 m^2-9}}{2 m}=\sqrt{91} \sqrt{25 m^2-9} \\ & \Rightarrow 2 m=1 \Rightarrow m=\frac{1}{2} \end{aligned} }

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Anam Khan

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