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  • The slopes of the common tangents to the hyperbola <img alt="\mathrm{\frac{x^2}{9}-\frac{y^2}{16}=1 \quad \text { and } \quad \frac{y^2}{9}-\frac{x^2}{16}=1 \text { are }}" src="https://entran

The slopes of the common tangents to the hyperbola \mathrm{\frac{x^2}{9}-\frac{y^2}{16}=1 \quad \text { and } \quad \frac{y^2}{9}-\frac{x^2}{16}=1 \text { are }}

Option: 1

- 2, 2


Option: 2

- 1, 1


Option: 3

1, 2


Option: 4

2, 1


Answers (1)

best_answer

Given hyperbola are \mathrm{\frac{x^2}{9}-\frac{y^2}{16}=1\ \ \ \ .....(i)\ and\ \frac{y^2}{9}-\frac{x^2}{16}=1\ \ \ .......{(ii)}}
Any tangent to \mathrm{(i)} having slope \mathrm{m} is \mathrm{y=m x \pm \sqrt{9 m^2-16}\ \ \ .........(iii)}
Putting in (ii), we get, \mathrm{16\left[m x \pm \sqrt{9 m^2-16}\right]^2-9 x^2=144}
\begin{aligned} &\mathrm{ \Rightarrow \quad\left(16 m^2-9\right) x^2 \pm 32 m\left(\sqrt{9 m^2-16}\right) x+144 m^2-256-144=0} \\ &\mathrm{ \Rightarrow\left(16 m^2-9\right) x^2 \pm 32 m\left(\sqrt{9 m^2-16}\right) x+\left(144 m^2-400\right)=0\ \ \ \ .........(iv)} \end{aligned}
If \mathrm{(iii)} is a tangent to \mathrm{(ii)}, then the roots of \mathrm{(iv)} are real and equal.
\mathrm{\begin{aligned} &\mathrm{ \therefore \text { Discriminant }=0 ; \quad 32 \times 32 m^2\left(9 m^2-16\right)=4\left(16 m^2-9\right)\left(144 m^2-400\right)=} \\ &\mathrm{ 64\left(16 m^2-9\right)\left(9 m^2-25\right)} \end{aligned}}
\begin{aligned} &\mathrm{ \Rightarrow \quad 16 m^2\left(9 m^2-16\right)=\left(16 m^2-9\right)\left(9 m^2-25\right) \Rightarrow 144 m^4-256 m^2=144 m^4-481 m^2+225} \\ &\mathrm{ \Rightarrow \quad 225 m^2=225 \Rightarrow m^2=1 \Rightarrow m= \pm 1} \end{aligned}

Posted by

Devendra Khairwa

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