JEE Main 2019 — Hyperbola Question with Solution

The slope of a line $ax+by+c=0$ is $m=-\frac{a}{b}.$

Thus, the slope of the line $x-y=2$ is $m=-\frac{1}{\left(-1\right)}=1$ and we know that if two lines are parallel then their slopes are equal, hence the slope of the tangent is also $m=1.$

The given hyperbola is $\frac{x^2}{5}-\frac{y^2}{4}=1$

The equation of the tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having slope $m$ is $y=mx\pm \sqrt{a^2{m}^2-b^2}.$

Hence, the equation of its tangent having slope $1$ is $y=x\pm \sqrt{5\times 1^2-4}$

$\Rightarrow y=x\pm 1$

Thus, the tangents are $x-y+1=0 \& x-y-1=0.$