A few pre-calc guidance needed

• January 9th 2008, 04:07 PM
supercheddarcheese
A few pre-calc guidance needed
Hi again. I have a few problems which I feel like I'm right at the finish line but I just can't cross...They are the following:

FIRSTLY

Let a and b be real numbers.

f(x)=a sin(x)+ b ∛x+4

If f(log(log(10,3))=5,

what is the value of f(log(log3))?

For this what i did was evaluate the log of 10 base 3, and then evaluate that...anyways, I evaluated eerything and then got the sin and 3rd roots of it, or x. I got the following:

.0056a+.685b=1 or f(.3214)=5

Whats next? I can evaluate log log 3 to get -.3214, so does that mean the answer is just -5? thanks..

SECOND

The graphs of (x/4)+(y/3)=1 and ((x^2)/16)+((y^2)/9)=1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3

So I graphed them(isolated y for firstequation) and I found a circle being interescted in two points. I found A to be (0,3) and B to be (4,0). Now, point c obviously has to be in the northeastern part of the ellipse within the range of the line. I have no idea how to find it and get an area of 3 though. Maybe use the midpoint? (2,1.5)?

Thanks

This is basically it for now. There are two other ones that I've solved but I'm not too sure about.
• January 9th 2008, 09:04 PM
earboth
Quote:

Originally Posted by supercheddarcheese
SECOND

The graphs of (x/4)+(y/3)=1 and ((x^2)/16)+((y^2)/9)=1 intersect at points A and B. Find the coordinates of a third point C on the graph of the second curve such that the area of triangle ABC is equal to 3

... I found A to be (0,3) and B to be (4,0). ........Correct

Hello,

I've attached a sketch of the ellips and the line.

Consider the distance between (0, 3) and (4, 0) to be the base of the triangle you are looking for. The base has a length of 5. Thus you need the height of the triangle so that the area is 3:

$A = \frac12 \cdot 5 \cdot h = 3~\iff~ h = \frac65$

Therefore you need a parallel to the given line with the distance of $\frac65$. Calculate the coordinates of a point on the y-axis which has a distance of $\frac65$ from the given line:
C(0, a)

Use the distance formula:

$\pm \frac65=\frac{3 \cdot 0 + 4 \cdot a -12}{\sqrt{3^2+4^2}} =\frac{4a-12}{5}~\implies~a = \frac92~\vee~a=\frac32$

Only $a = \frac32$ is a valid value (otherwise the line will not intersect with the ellipse)

The parallel to the given line has the equation: $y = -\frac34 \cdot x + \frac32$

To get the coordinates of the points R and V calculate the intersection points of the ellipse and the parallel:

$\frac{x^2}{16}+\frac{\left( -\frac34 \cdot x + \frac32 \right)^2}{9}=1~\implies~x = 1-\sqrt{7}~\vee~x=1+\sqrt{7}$

Plug in these x-values into the equation of the parallel to get the y-coordinates of the points R and V.