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

Mar 31, 2018

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

#### Explanation:

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

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

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

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

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

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

i.e. $m = \pm 1$

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

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

Mar 31, 2018

$\text{Slope of tangent } = \frac{\mathrm{dy}}{\mathrm{dx}}$

Therefore, slope of tangent of x^2−y^2=1 => dx^2/dx−dy^2/dx=d1/dx

$\implies 2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$

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

$\implies m = \frac{x}{y}$