Equation tangent to a line

girlie01

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Jan 18, 2021
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Question: Find the equation tangent to the line y= 4 at points (6,4) and also tangent at the line x= 10 at point (10,0).

Hi, I need help with this one. I'm completely new to the concept behind it so I'm fairly confused about how to go about it.

I know I should be using this: y - y1 = m(x-x1)
For the first one: y1 = 4, x1 = 6
So, y - 4 = m (x - 6)
This is where I'm stuck cause I think I'm missing something. What do I do with the y =4, as well as the x=10? I do not know how to connect them together. Or I should consider this: y2 -y1/x2-x1 = 0 - 4 / 10 - 6 = -1. So m = -1. The solution: y - 4 = -1 (x-6).
y - 4 = -x + 6
y = -x + 10.
I'm unsure of how to go through the problem or if I need a derivative for this. Your help will be very much appreciated. Thank youuu
 

Harry_the_cat

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Mar 16, 2016
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Can you double check that you have stated the question exactly as it was given to you? Sounds like you might have left something out.
 

lev888

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Jan 16, 2018
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Question: Find the equation tangent to the line y= 4 at points (6,4) and also tangent at the line x= 10 at point (10,0).

Hi, I need help with this one. I'm completely new to the concept behind it so I'm fairly confused about how to go about it.

I know I should be using this: y - y1 = m(x-x1)
For the first one: y1 = 4, x1 = 6
So, y - 4 = m (x - 6)
This is where I'm stuck cause I think I'm missing something. What do I do with the y =4, as well as the x=10? I do not know how to connect them together. Or I should consider this: y2 -y1/x2-x1 = 0 - 4 / 10 - 6 = -1. So m = -1. The solution: y - 4 = -1 (x-6).
y - 4 = -x + 6
y = -x + 10.
I'm unsure of how to go through the problem or if I need a derivative for this. Your help will be very much appreciated. Thank youuu
Equation of what?
 

HallsofIvy

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Jan 27, 2012
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7,768
Do you understand what a "tangent" is? The tangent to a graph at a point is a straight line that passes through that point and has the same slope (derivative) there.

Do you understand what "y= 3" means? Since y is always 3 any (x, y) point is (x, 3). The graph is a horizontal straight line! Any "tangent line" is that horizontal straight line!
 

girlie01

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Jan 18, 2021
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Can you double check that you have stated the question exactly as it was given to you? Sounds like you might have left something out.
I am sorry for the very late reply. But yes, this is what is exactly stated. I have double-checked it.
 

girlie01

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Do you understand what a "tangent" is? The tangent to a graph at a point is a straight line that passes through that point and has the same slope (derivative) there.

Do you understand what "y= 3" means? Since y is always 3 any (x, y) point is (x, 3). The graph is a horizontal straight line! Any "tangent line" is that horizontal straight line!
Ohh, so you mean to say that the equation is based on that horizontal line because that is exactly the tangent line. I have an idea of it now. Thank you
 

Dr.Peterson

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Question: Find the equation [of what?] tangent to the line y= 4 at points (6,4) and also tangent at the line x= 10 at point (10,0).
Clearly you are not being asked for the equation of a line tangent to both of these lines; that is impossible.

Possibly the problem is supposed to be this:

Find the equation of a circle tangent to the line y= 4 at point (6,4) and also tangent at the line x= 10 at point (10,0).​

Can you try solving that? What two lines does the center of this circle have to lie on?
 

girlie01

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Jan 18, 2021
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Clearly you are not being asked for the equation of a line tangent to both of these lines; that is impossible.

Possibly the problem is supposed to be this:

Find the equation of a circle tangent to the line y= 4 at point (6,4) and also tangent at the line x= 10 at point (10,0).​

Can you try solving that? What two lines does the center of this circle have to lie on?
This makes so much sense. So, I would be using the formula: (x-h)^2 + (y-k)^2 = r^2. I should plug in the points, however, I'm not quite sure what to substitute on r^2. Or, should I use the distance formula which is: square root of (x2-x1) + (y2-y1) to find r first?
 

Dr.Peterson

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This makes so much sense. So, I would be using the formula: (x-h)^2 + (y-k)^2 = r^2. I should plug in the points, however, I'm not quite sure what to substitute on r^2. Or, should I use the distance formula which is: square root of (x2-x1) + (y2-y1) to find r first?
No, you don't know the center or radius, so you have to think first! (You might want to sketch a graph.)

Answer my question: If a circle is tangent to y=4 at (6,4), what line must the center be on?

Also, do you have good reason to think the question is about a circle? That was just a guess. What topic have you been learning?
 

girlie01

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Jan 18, 2021
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No, you don't know the center or radius, so you have to think first! (You might want to sketch a graph.)

Answer my question: If a circle is tangent to y=4 at (6,4), what line must the center be on?

Also, do you have good reason to think the question is about a circle? That was just a guess. What topic have you been learning?
I'm sorry for the late reply. The center must be on (10,4). And the points are four units away from the center. Our topic is conic sections, so yes it must be a circle. With this information, the equation of the line is (x-10)^2 + (y-4)^2 = 16. I'm not sure if I did this correctly. Thank you
 

jonah2.0

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Apr 29, 2014
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Beer induced reaction follows.
I'm sorry for the late reply. The center must be on (10,4). And the points are four units away from the center. Our topic is conic sections, so yes it must be a circle. With this information, the equation of the line is (x-10)^2 + (y-4)^2 = 16. I'm not sure if I did this correctly. Thank you
If Dr.Peterson's conjecture is correct, I'd say you didn't do that correctly.
I say try again and try it again this time with visual feedback.
https://www.desmos.com/calculator/rprrvewzus
 
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