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

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Question

(A) The given hyperbola is $4x^2 - 5y^2 = 20$,which can be written as $\frac{x^2}{5} - \frac{y^2}{4} = 1$. Comparing this with $\frac{x^2}{a^2} - \fra

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$x - y + 1 = 0$

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(A) The given hyperbola is $4x^2 - 5y^2 = 20$,which can be written as $\frac{x^2}{5} - \frac{y^2}{4} = 1$. Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 5$ and $b^2 = 4$. The given line is $x - y = 2$,which can be written as $y = x - 2$. The slope of this line is $m = 1$. The equation of a tangent to the hyperbola with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$. Substituting $m = 1$,$a^2 = 5$,and $b^2 = 4$,we get $y = 1(x) \pm \sqrt{5(1)^2 - 4} = x \pm \sqrt{5 - 4} = x \pm 1$. Thus,the tangents are $y = x + 1$ or $y = x - 1$,which can be written as $x - y + 1 = 0$ or $x - y - 1 = 0$. Comparing with the given options,$x - y + 1 = 0$ is the correct choice.

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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