1. ## graph equations

the curve, C has the equation:
$4(x+1)^2 - (y-1)^2 =1$

Find the greatest value of k, where k is a positive integer, for which the curve $y=ln(x+k)$ cuts C at only one point.

what i tried to do is:
$4(x+1)^2 - (y-1)^2 =1$
$4x^2 + 8x - y^2 +2y +2=0$
By substitution,
$4x^2+8x-[ln(x+k)]^2+2[ln(x+k)]+2=0
$

and then i'm stuck because of $[ln(x+k)]^2$.
is the approach wrong or did i make a mistake somewhere?

2. Hello !

You have two curves, $4(x+1)^2 - (y-1)^2 =1$ and $y=\ln{(x+k)}$. You wish to find the greatest number $k$ for which the curves intersect only once.

Let us substitute the second curve into the first one (since the $y$ values are presumed equal as there is intersection) :

$4(x+1)^2 - (\ln{(x+k)}-1)^2 =1$

Now let us reformulate the problem : you wish to find the greatest number $k$ for which this equation has only one solution for $x$ (equivalent to the statement "the curves only intersect once").

Expand this :

$4(x+1)^2 - (\ln{(x+k)}-1)^2 =1$

$4(x^2 + 2x + 1) - (\ln{(x+k)}^2 - 2 \ln{(x+k)} + 1) = 1$

Ok, keep going :

$4x^2 + 8x + 4 - \ln{(x+k)}^2 + 2 \ln{(x+k)} - 1 = 1$

Let's try to set this as a quadratic equation to use the discriminant :

$4x^2 + 8x - \ln{(x+k)}^2 + 2 \ln{(x+k)} + 2 = 0$

Consider everything after the $8x$ as the constant coefficient (yeah, who said we couldn't put $x$'s in it ?)

You know that if the discriminant of a quadratic equation is equal to zero, then there is only one real solution. So, let us express the discriminant :

$\Delta = b^2 - 4ac = 64 - 16 \times (\ln{(x+k)}^2 + 2 \ln{(x+k)} + 2)$

You want it to be equal to zero, so attempt to solve the following for $k$ (yes, now we care about $k$ because we are looking for the value of $k$ that makes the discriminant equal to zero) :

$64 - 16 \times (\ln{(x+k)}^2 + 2 \ln{(x+k)} + 2) = 0$

Simplify :

$64 = 16 \times (\ln{(x+k)}^2 + 2 \ln{(x+k)} + 2)$

Again :

$4 = \ln{(x+k)}^2 + 2 \ln{(x+k)} + 2$

$0 = \ln{(x+k)}^2 + 2 \ln{(x+k)} - 2$

$\ln{(x+k)}^2 + 2 \ln{(x+k)} - 2 = 0$

Cool, a straightforward quadratic equation ! Here we are using an advanced method : we are going to solve for $\ln{(x + k)}$, and this will serve as an intermediate result to determine which values of $k$ make the discriminant null. Get the discriminant first, for ease :

$\Delta = 2^2 - 4 \times (-2) = 4 + 4 \times 2 = 12$

$\ln{(x+k)} = \frac{-4 \pm \sqrt{12}}{2}$

So if some $k$ makes the two curves intersect, then we have $\ln{(x+k)} = \frac{-4 \pm \sqrt{12}}{2}$ for the $x$ where the curves intersect (discard the negative value since a logarithm is always positive)

Now you can apply the same method but solving for $x$ instead, and then merge the results together to find a condition on $k$ allowing you to find the greatest $k$ that satisfies your problem.

PS : I think I got way too complicated here, I don't even know if this actually works, there surely is a simpler way to do this, what have you been doing last with your class ? (or elsewhere)

3. Never mind what I wrote before, it's just wrong (fail). However, in your post I see a mistake, it should be -2 and not +2 in the two last lines.

I don't have time now, so here is a little hint : try to use the discriminant to avoid messing with the logarithms, and remember the graphic interpretation of the quadratic discriminant.

Although 3 unknowns make this really harsh.