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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) $y=3x+5, y=3x+1$

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Please clarify the question. So you seek a line that is parallel to $3x-y=-6 \Longleftrightarrow y = 3x +6$? Then the only candidates could be lines with the same gradient, i.e. those with a $3x$-term... –  Sh4pe Oct 26 '12 at 12:24
    
Sorry, wanted to write "e.g. those with a $3x$-term". –  Sh4pe Oct 26 '12 at 12:30
    
As far as I can tell, there is something wrong with this problem. You are looking for tangent lines with slow $3$, but only one tangent line has slope $3$. –  Thomas Andrews Oct 26 '12 at 13:02
    
Why is this tagged as algebraic geometry question? It should be analytic geometry instead. –  Manos Oct 26 '12 at 16:10
    
That is all the info. i was given to do the question. My apologies for the incorrect tag but i was rushing when i put this up. –  Serena Oct 27 '12 at 1:12
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3 Answers 3

A line being tangent to a curve means that it "touches it" i.e. there is a shared point of the line and the curve, and at that point the gradients are the same. Two lines being parallel means that they have the same gradient.

What you need to do is find the general equation of any line parallel to the given one, then find the points of intersection of this general parallel line with the given quadratic curve. The lines that are tangent to the curve will satisfy the condition that, at the points where the lines intersect the curve, the gradient of the line will be the same as the gradient of the curve.

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Thanks you very much –  Serena Oct 27 '12 at 20:20
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As said above, the slope of the required line is 3 but there will be only one line which is tangent to the curve. The given curve is a parabola and there is only one tangent of a given slope touching it.

$y=3 +x^2$

$dy/dx = 2x$

Since the slope of the line is 3

$dy/dx=2x=3$

Therefore, $x=1.5 $ and $y=3+(1.5)^2 =5.25$

Therefore the equation of given line will be $y=3x+0.75$

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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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thank you for your help, much appreciated –  Serena Oct 27 '12 at 1:16
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