<s>[INST] Finding the equations of two lines that are tangent to the curve and parallel to a line. I have no idea how to do this as i have been absent from class for medical reasons.
the curve $y=3+x^2$
and the line $3x-y=-6$
The answer is a possibility of the following:
a) $ y=3x, y=3x+4$
b) $ y=12x-15, y=12x+17$
c) $ y=-3x+9, y=-3x-16$
d) $y=12x-13, y=12+19$
e) $y=-12x-16, y=-12x+8$
f) $y=3x+5, y=3x+1$
 [/INST] The equation of any line parallel to $3x-y=-6$ can be written as $3x-y=k$
Now, let this line touches the curve $y=3+x^2$ at $(a,b).$
So, $3a-b=k$ or $b=3a-k$  and $b=3+a^2$
Comparing the values of $b$ we get, $3a-k=3+a^2,$ or $a^2-3a+k+3=0$
Now, this is a quadratic equation in $a,$ for the tangency both the root should be same, i.e, discriminant$(-3)^2-4\cdot 1\cdot(k+3)$ must be $0\implies 9-4k-12=0,k=-\frac 3 4$ 
So, the equation of the tangent becomes  $3x-y=-\frac 3 4,$ or $12x-4y=-3$
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