1.

Find the equation of tangents to hyperbola `x^(2)-y^(2)-4x-2y=0` having slope 2.

Answer» We have hyperbola
`x^(2)-y^(2)-4x-2y=0`
`"or "(x-2)^(2)-(y+1)^(2)=3`
`"or "((x-2)^(2))/(3)-((y+1)^(2))/(3)=1" (1)"`
Equation of tangents of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` having slope m is given by
`y=mxpm sqrt(a^(2)m^(2)-b^(2))`
So, equation of tangents to hyperbola (1) having slope 2 is given by
`y+1=2(x-2)pmsqrt(3xx4-3)`
`"or "y+1=2x-4pm3`
Therefore, the equations of tangents to given hyperbola are `2x-y-2=0` and `2x-y-8=0.`


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