An Unknown Conic and Equation of Tangent Line HELP

  • #1
An Unknown Conic and Equation of Tangent Line **HELP**

Homework Statement


(a) Find the positive y-coordinate of the point on the conic whose x-coordinate is 27cos[tex]\theta[/tex], where 0<[tex]\theta[/tex]<[tex]\pi[/tex]/2
(b) The slope of the tangent line to the conic through the point (27cos[tex]\theta[/tex], y1) is -1/27cot[tex]\theta[/tex], where y1 is your solution in part (a). Find the equation of the tangent line to the conic through the point (27cos[tex]\theta[/tex], y1)
(c) Write an expression for S where S is the sum of the x- and y-intercepts of the tangent line that you computed in part (b)
(d) If [tex]\theta[/tex] = tan-1(1/3), show that the value of S is 103/2

Homework Equations



x2 / 272 + y2 = 1


The Attempt at a Solution


okay sooo...
because it has the + sign I think it is an Ellipse with the general equation of:
(x-h)2/a2 + (y-k)2/b2 = 1

BUT I graphed it using my calculator and it gave me an almost straight line, to do this I rearranged the equation to give:
y = 1 - [tex]\sqrt{x^2/27^2}[/tex]

for part (d)[tex]\theta[/tex] = tan-1(1/3) then tan[tex]\theta[/tex] = 1/3
YA...I'm stuck:cry:
PLEASE HELP Guys:blushing:
 

Answers and Replies

  • #2


lol does no one want to help with this one or what?!:smile:
 
  • #3
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Is x^2/27^2 + y^2 = 1 the equation of the conic?

I don't think you're giving us all the information.
 
  • #4


Yes, Mark that is the equation of the conic given. The only part of the question I did not type out is the first part:
You are living in a time just before Newton ad Leibniz. One night as you sleep, you have dream involving the conic:
x2/272 + y2 = 1

I didn't think it mattered lol
 
  • #5
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Yeah, it matters a lot!
Since you're living before Newton and Leibniz, there is no calculus. The equation is that of an ellipse whose center is at (0, 0) and whose vertices are at (+/-27, 0) and (0, +/-1).

If you solve x2/272 + y2 = 1 for y, you don't get what you did. Check your work and try again to find y_1.
You are given the slope of the tangent line at (27 cos(theta), y_1). Hopefully you can find the equation of this line once you know a point on the line and its slope.
 
  • #6


Yeah, it matters a lot!
Since you're living before Newton and Leibniz, there is no calculus. The equation is that of an ellipse whose center is at (0, 0) and whose vertices are at (+/-27, 0) and (0, +/-1).

If you solve x2/272 + y2 = 1 for y, you don't get what you did. Check your work and try again to find y_1.
You are given the slope of the tangent line at (27 cos(theta), y_1). Hopefully you can find the equation of this line once you know a point on the line and its slope.

hi mark, I just cannot see how my y is wrong?! Here's what I did:
(x^2/27^2) + y^2 = 1
y^2 = 1 - (x^2/27^2)
y =sqrt{ 1 - (x^2/27^2)}

Would I now plug in 27cos(theta) for my x? But what is theta? All I know about the tangent line is y = (-1/27cot(theta))x + b --> to find b I habe to plug in the x = 27cos(theta) and the y that I haven't found right?
thanks already
 
  • #7
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Well, pal.. i dont find a reason to make this thing so complicated, when it isnt! :).

Go a bit systematically, you have been given the equation of the ellipse which is,
[tex]\frac{x^2}{27^2} +\frac{y^2}{1^2}=~1[/tex]

Now substitue, the value of x in the above given equation,
thus, you have,
[tex]\frac{(27cos\theta)^2}{27^2} +\frac{y^2}{1^2}=~1[/tex]
now, cancelling [tex]27^2[/tex] from the numerator and the denomintor, you get,
[tex]y^2=~1-cos^2\theta[/tex]
Therefore, [tex]y=sin\theta[/tex]

Now try solving the problem further, buddy!
 
  • #8


Well, pal.. i dont find a reason to make this thing so complicated, when it isnt! :).

Go a bit systematically, you have been given the equation of the ellipse which is,
[tex]\frac{x^2}{27^2} +\frac{y^2}{1^2}=~1[/tex]

Now substitue, the value of x in the above given equation,
thus, you have,
[tex]\frac{(27cos\theta)^2}{27^2} +\frac{y^2}{1^2}=~1[/tex]
now, cancelling [tex]27^2[/tex] from the numerator and the denomintor, you get,
[tex]y^2=~1-cos^2\theta[/tex]
Therefore, [tex]y=sin\theta[/tex]

Now try solving the problem further, buddy!

hey you, thanks alot! So when I plug it into y=mx+b I get:
sin(theta)=(-1/cot(theta))cos(theta) + b is this the answer to part (b)??
what is (c) asking for here?!
calculus is another language! :(
 
  • #9
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Rather, the solution to (b) would be,
[tex]sin\theta=~\frac{-1}{27cot\theta}.27cos\theta +~b[/tex]
Therefore,
[tex]sin\theta=~\frac{-sin\theta}{27cos\theta}.27cos\theta +~b[/tex]

Thus,
27cos\theta gets cancelled and the value of b turns out to be [tex]2sin\theta[/tex]
Therefore, the equation would be,
[tex]27ycos\theta +~xsin\theta=~2sin\theta [/tex]
(Do try it once again!)
 
  • #10


Rather, the solution to (b) would be,
[tex]sin\theta=~\frac{-1}{27cot\theta}.27cos\theta +~b[/tex]
Therefore,
[tex]sin\theta=~\frac{-sin\theta}{27cos\theta}.27cos\theta +~b[/tex]

Thus,
27cos\theta gets cancelled and the value of b turns out to be [tex]2sin\theta[/tex]
Therefore, the equation would be,
[tex]27ycos\theta +~xsin\theta=~2sin\theta [/tex]
(Do try it once again!)

Oh I see what you did there, you converted cot to sin/cos...genius :D
so now this is the equation for the tangent line, awesome...next problem, what am I supposed to do for part (c), what is the question asking for? the Sum of the x and y intercepts???
 
  • #11
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Pal, I've helped you a lot with this, you need to have your own input over it, give it a try!
 
  • #12


Pal, I've helped you a lot with this, you need to have your own input over it, give it a try!

:cry:could you maybe just reword it or something? sorry I know you've helped me loads but it seems to be hard for me understand what the question is asking for:frown: I'm german :redface:
 
  • #13
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Well, please check the question once! I'm facing a petty problem..
 
  • #14


Uhm, okay so I just wrote the entire question out again and tried to find out what 'the sum of the x and y intercepts' means but I cannot get any further with my problem :(
 
  • #15
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7,294


You have the equation of the tangent line.
  1. Find the x-intercept of this line.
  2. Find the y-intercept of this line.
  3. Add the two intercepts together.
 
  • #16


You have the equation of the tangent line.
  1. Find the x-intercept of this line.
  2. Find the y-intercept of this line.
  3. Add the two intercepts together.


OKAY so:
27ycos[tex]\theta[/tex] = -(sin[tex]\theta[/tex])x + 2sin[tex]\theta[/tex]

x-intercept:
let y=0
0 = -sin[tex]\theta[/tex]x + 2sin[tex]\theta[/tex]
(sin[tex]\theta[/tex])x + 2sin[tex]\theta[/tex]
x = (2sin[tex]\theta[/tex])/sin[tex]\theta[/tex]


y-intercept:
let x=0
27ycos[tex]\theta[/tex] = 0 + 2sin[tex]\theta[/tex]
27ycos[tex]\theta[/tex] = 2sin[tex]\theta[/tex]
y = 2sin[tex]\theta[/tex]/27cos[tex]\theta[/tex]


Sum of x- & y-intercepts:
S = 2sin[tex]\theta[/tex]/sin[tex]\theta[/tex] + 2sin[tex]\theta[/tex]/27cos[tex]\theta[/tex]
S = 2 + (2/27)tan[tex]\theta[/tex]


for part (d):
[tex]\theta[/tex] = tan-11/3
I will now plug this in to S from part (c):
S = 2 + (2/27)tan(tan-11/3
S = 2.025
WRONG --> the answer is supposed to be 102/3 which is 31.62 ????
What is wrong here? I don't understand??:confused:
 
  • #17
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You are so wrong for part c and d, wait for Prof Sandeep/Sarkar to post up the solutions in clew.
 
  • #18


You are so wrong for part c and d, wait for Prof Sandeep/Sarkar to post up the solutions in clew.

lmao Okay, MrUnknown who is either Justin or someone who knows Justin...that would be alittle late dont you think?...why dont you tell me the right answer then?
 
Last edited:
  • #19
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Yes, thats what Anna, I could not get to [tex]10^{3/2}[/tex], which is why i asked you check the question! As of finding the intercepts, all looks good to me!
 
  • #20


Yes, thats what Anna, I could not get to [tex]10^{3/2}[/tex], which is why i asked you check the question! As of finding the intercepts, all looks good to me!

o wow that sucks...a lot i just checked, i typed everything right...no mistakes :(
 
  • #21
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a) y1 = sin [itex]\theta[/itex] (You have that in post 7.)

b) Tangent line equation at the given point:
[tex]y = \frac{-cos \theta}{27 sin \theta}x + \frac{1}{sin \theta}[/tex]
I don't think you ever got this, and I think psykatic's work is in error, but only because of how you provided the information all the way back in your first post. You have
The slope of the tangent line to the conic through the point (27cos, y1) is -1/27cot [itex]\theta [/itex]
I believe that the slope that is given is -1/27 * cot [itex]\theta [/itex]. I believe that Psykatic interpreted what you wrote as -1/(27 * cot [itex]\theta [/itex]), which is different. In the first expression cot([itex]\theta [/itex]) is in the numerator; in the latter expression it is in the denominator.

c) y-intercept: 1/sin [itex]\theta [/itex]
x-intercept: 27/cos [itex]\theta [/itex]
S = 1/sin [itex]\theta [/itex] + 27/cos [itex]\theta [/itex]

d) [itex]\theta [/itex] = tan-1(1/3)
Since you're going to need the sine and cosine of this angle, it's helpful to draw a right triangle whose base is length 3 and whose altitude is 1. The angle is [itex]\theta [/itex].
From this triangle, it's fairly easy to determine that sin([itex]\theta [/itex]) is 1/sqrt(10) and that cos([itex]\theta [/itex]) is 3/sqrt(10).

So S = 1/sin([itex]\theta [/itex]) + 27/cos([itex]\theta [/itex])
= sqrt(10) + 27 sqrt(10)/3 = sqrt(10) + 9sqrt(10) = 10sqrt(10) = 103/2.

This is post #21. It shouldn't take this many posts to work through this problem. I didn't show all of my work here, but I did the entire problem on about a half a page of legal-size paper. If you're rusty with the algebra and/or trig, I would strongly encourage you to put in some time at reviewing that material, otherwise you're going to keep struggling with calculus.
 
  • #22
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Well, I spent 2 full size papers working on it! And I had so many doubts rushing one after the another, and that was because of just one wrong interpretation, gawd!

Mark44, I did take the slop to be, [tex]\frac{-1}{27cot\theta}[/tex], members should start using LaTex codes now, understanding the question becomes the first problem rather than trying its solution, :)!

Now, that I know the exact thing, I got the answer as well!

P.S: Anna, I'd told you to check the question, :rolleyes:!
 

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