The equation of a tangent to the hyperbola 4x | vedclass

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Question

Hyperbola is $\frac{{{x^2}}}{5} - \frac{{{y^2}}}{4} = 1$

Equation of its tangent in slop from is $y = mx \pm \sqrt {5{m^2} - 4} $

Henc

Vedclass · Accepted Answer

$x -y + 1 = 0$

Vedclass · Answer

Hyperbola is $\frac{{{x^2}}}{5} - \frac{{{y^2}}}{4} = 1$

Equation of its tangent in slop from is $y = mx \pm \sqrt {5{m^2} - 4} $

Hence tangent with slope $1$ is $y = x \pm 1$

The equation of a tangent to the hyperbola $4x^2 -5y^2 = 20$ parallel to the line $x -y = 2$ is

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