Calculus Problem Solving- help!?

Question

Consider a pt in the first quadrant and on the parabola y=4-x^2. If we draw a tangent at this point, the tangent , x axis and y axis form a triangle. Find the coordinates of the point on the curve so that the area of this triangle is 6.5 units.

(I got 1/13, Im not sure though. And they say the other answer is 2705/676. I dont know how to get it)

I also end up having the equation. I dont know how to get the coordinates

Answer 1

Both the answers, the one that you are proposing and the one that they are giving to you, DO NOT LIE ON THE GIVEN PARABOLA
so i suspect a typo error ??
oyherwise the solution will be as follows,
tangent at (a,b) is (y+b)/2 = 4 - ax
ie, y + 2ax = 8-b
intercepts on x and y are (8-b)/2a and 8-b
area will be (8-b)^2/4a
now, (a,b) is on the curve so
4-b = a^2
therefore area = (4+a^2)^2/4a = 13/2
that is, a^4 + 8a^2 + 16 = 26a
or, a4 + 8a2 -26a + 16 = 0
so a = .87 or 1.4945
b = 4-a^2 so b is 3.2431 or 1.7665
both the points (0.87, 3.243) and (1.4945, 1.7665) satisfy all the condition and are in the first quadrant
In my openion these are valid solutions...
and ... you got two more answers! all different!!

Answer 2

Ya know, I actually spent an hour on the other version of this question last evening before you deleted it. That's fine, though, because I didn't post an answer because I got stuck at the same point as everyone else.

I decided to watch everyone else's answers...and I'm glad to know they all got stuck too.

So today, I think I'll try looking at it in reverse.

First off, let's assume that their y-coordinate, y = 2705/676, is correct. If so, then the x coordinate would be:

2705/676 = 4 - x²
x² = 4 - 2705/676 = 2704/676 - 2705/676 = -1/676

Uh-oh. Problem there...can't take the square root of a negative number.... but it's interesting that, if you take away the minus sign, you'd get your answer, 1/13.

Which suggests that the problem should have been:

y = 4 + x²

But in that case, you wouldn't get a negative slope, and therefore you wouldn't get a triangle.

What if we assume that the 4 is wrong? I.e. plug the point (1/13, 2705/676) into y = k - x² and find k:

2705/676 = k - (1/13)² = k - 1/169

k = 2705/676 + 4/676 = 2709/676 = 4.0073964497

That's not likely to help.

Can you please post your work to show how you got 1/13?

Answer 3

OK - this is a little long and probably has some error. See the following, please :

Step 1 : Any line that is acting as a tangent to the given parabola has a slope equal to derivative of the curve defining the parabola. So, if the line that is acting as a tangent were to be represented as Y=MX+C, we have, the curve equation of parabola as
y=4-x^2
So, dy/dx = -2x
So, M=-2
So, the line is defined as Y=-2X+C

Step 2 : Such a line will intersect x - axis at Y=0, so will intersect x-axis at X=C/2 (put Y=0 in the equation for the line above)

Step 3 : The same line will intersect Y axis at X=0, so will intersect at Y=C

Step 4 : Hence area of the triangle enclosed within the two axes and the line is 1/2(C/2)(C) = C^2/4 = given 6.5. Therefore, C = sqrt(26).

Step 5 : If the point of tangency had a co-ordinate of (x1, y1) then, this point will lie on the tangent line Y=-2X+sqrt(26) and on the parabola y=4-x^2. So, the following two relations will be satisfied :

y1 = -2(x1) +sqrt(26)
y1 = 4 - x1^2

Equating,
4-x1^2=-2(x1)+sqrt(26)
Or, in simpler terms, x1^2 -2x1+(sqrt(26)-4) = 0

Step 6 : This quadratic has no solution. The maximum area that may be supported by a triangle acting as a tangent to the given parabola is 9/4

Please check the problem, and do let me know if I am committing an error somewhere....

Answer 4

y=4-x^2
slope of the tangent =dy/dx = - 2x.
Let the required point on the parabola be (a,b).
then: b=4-a^2
So slope of the tangent at (a,b) = -2a.
So the Eq. of the tangent is y-b=-2a(x-a).
It cuts the X-axis where y=0, x=a+b/(2a)
=(b+2a^2)/(2a)= (4+a^2)/(2a)
It cuts the Y-axis where x=0, y=(b+2a^2)
= (4+a^2)
So the area of the triangle formed by the sides which are the X-axis, Y-axis and the tangent = (1/2)(4+a^2)^2/(2a)
So (1/2)(4+a^2)^2/(2a) = 6.5
(4+a^2)^2 = 26a.
Or a^4 +8a^2- 26a +16=0

Answer 5

y = 4 - x^2

First off, since tangents are related to derivatives, let's take the derivative of that above function.

y' = -2x

For some unknown value t

m = -2t

Therefore, our unknown slope m = -2t

The base and height of the triangle is related to the slope, because the height represents the y-intercept of the line, and the base represents the x-intercept.

Let's get the equation of the line. Note that if x = t, then
y = 4 - t^2, so the equation of the line is

(y - (4 - t^2)) / (x - t) = -2t

(y - 4 + t^2) / (x - t) = -2t
(y - 4 + t^2) = -2t(x - t)
(y - 4 + t^2) = -2tx + 2t^2
y = -2tx + 2t^2 + 4 - t^2
y = (-2t)x + (t^2 + 4)

Now that we have the equation of our line, our y-intercept is
t^2 + 4 (which represents the height of our triangle).
To get our x-intercept, we make y = 0.

0 = (-2t)x + t^2 + 4
(-t^2 - 4)/(-2t) = x, OR
x = (t^2 + 4)/(2t), and this is our x-intercept.

The formula for the area of a triangle is (1/2)bh, so since
b = (t^2 + 4)/(2t) and h = t^2 + 4, it follows that

A = (1/2) [(t^2 + 4)/(2t)] [t^2 + 4]

However, we KNOW that A = 6.5, so

6.5 = (1/2) [(t^2 + 4)/(2t)] [t^2 + 4]

Merging the fractions, we have

6.5 = (1/(4t)) [t^2 + 4] [t^2 + 4]

Multiplying both sides by 4t,

(6.5)4t = [t^2 + 4][t^2 + 4]
26t = [t^2 + 4][t^2 + 4]

Expanding,

26t = t^4 + 8t^2 + 16
0 = t^4 + 8t^2 - 26t + 16

And I think I took a wrong step....

Answer 6

The "otheranswer" you give, 2705/676 is not in quadrant I
y = 4 - x^2
y' = -2x
y - b = -2a(x - a)
y - 4 + a^2 = -2ax + 2a^2
y = -2ax + a^2 + 4
y(0) = a^2 + 4
x = -(y - 4 - a^2)/2a
x(0) = (a^2 + 4)/2a
x(0)*y(0) = 13
(a^2 + 4)^2/2a = 13
a^4 + 8a^2 - 26a + 16 = 0

a ≈ 0.87, 1.4945

Answer 7

sin(a million/x)-x*cos(a million/x)(a million/x^2) i think of that's authentic utilising the product rule then the chain rule on the final section. "spinoff of x circumstances sin(a million/x) plus x circumstances the spinoff of sin(a million/x)"