For what values of $m$ $m$ is line $y = m x$ $y=mx$ tangent to the hyperbola $x^{2} - y^{2} = 1$ $x^2-y^2=1$ ?

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For what values of $m$ $m$ is line $y = m x$ $y=mx$ tangent to the hyperbola $x^{2} - y^{2} = 1$ $x^2-y^2=1$ ?

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Answer 1 · 2018-03-31T14:47:22

$m = \pm 1$ $m=+-1$ and tangents are $y = x$ $y=x$ and $y = - x$ $y=-x$

Explanation:

Put $y = m x$ $y=mx$ in te equation of hyperbola $x^{2} - y^{2} = 1$ $x^2-y^2=1$ , then

$x^{2} - m^{2} x^{2} = 1$ $x^2-m^2x^2=1$

or $x^{2} (1 - m^{2}) - 1 = 0$ $x^2(1-m^2)-1=0$

the values of $x$ $x$ will give points of intersection of $y = m x$ $y=mx$ and $x^{2} - y^{2} = 1$ $x^2-y^2=1$ . But as $y = m x$ $y=mx$ is a tangent, weshould get only one root, which would be wwhen discriminant is zero i.e.

$0^{2} - 4 \cdot (1 - m^{2}) \cdot (- 1) = 0$ $0^2-4*(1-m^2)*(-1)=0$

or $4 - 4 m^{2} = 0$ $4-4m^2=0$

i.e. $m = \pm 1$ $m=+-1$

and tangents are $y = x$ $y=x$ and $y = - x$ $y=-x$

graph{(x^2-y^2-1)(x+y)(x-y)=0 [-10, 10, -5, 5]}

Answer 2 · 2018-03-31T14:48:23

$Slope of tangent = \frac{d y}{d x}$ $"Slope of tangent " = dy/dx$

Therefore, slope of tangent of $x^{2} - y^{2} = 1 \Rightarrow \frac{{d x}^{2}}{d x} - \frac{{d y}^{2}}{d x} = d \frac{1}{d x}$ $x^2−y^2=1 => dx^2/dx−dy^2/dx=d1/dx$

$\Rightarrow 2 x - 2 y \frac{d y}{d x} = 0$ $=> 2x-2y dy/dx =0$

$\Rightarrow \frac{d y}{d x} = \frac{x}{y}$ $=> dy/dx =x/y$

For what values of m is line $y = m x$ $y=mx$ tangent to the hyperbola $x^{2} - y^{2} = 1$ $x^2−y^2=1$ ?

$\Rightarrow m = \frac{x}{y}$ $=> m =x/y$

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For what values of $m$ $m$ is line $y = m x$ $y=mx$ tangent to the hyperbola $x^{2} - y^{2} = 1$ $x^2-y^2=1$ ?

2 Answers

Explanation:

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For what values of $m$ $m$ is line $y = m x$ $y=mx$ tangent to the hyperbola $x^{2} - y^{2} = 1$ $x^2-y^2=1$ ?