# Thread:

1. ## 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.

2. 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.
Attached Thumbnails
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9327799081802368, "perplexity_flag": "head"}