Thread: Finding coordinates in hyperbolic-tangent is parallel?

I'm looking for other coordinates on the curve where the tangent is parallel to y=bx+3. I have solved for a, b, c and tangent equation.

Question.
(-3,a) is on curve y=5x-7/2x+10. The tangent at (-3,a) is parallel to y=bx+3 and cuts the y-axis at (0,c).

How do I get the other coordinates on the curve where the tangent is parallel to y=bx+3?
I’m sure I need to use the gradient perpendicular to the tangent (m=-1/4) and set it equal to y=5x-7/2x+10 or y`equation and solve for x?
Any help would be appreciated.


Workings
I have solved a,b,c (-5.5,4,6.5) and the tangent equation y=4x+6.5. But graphing the curve shows me there is 2 hyperbolas so I need the points on the curve for the other tangent equation.

1. st x = -3 into y=(5x-7)/(2x+10) to get -5.5=a ;. (-3,-5.5).
2. find y` of y=(5x-7)/(2x+10). Use quotient rule..vu`-uv`/v^2.
Then you get [5(2x+10)-2(5x-7)]/[(2x+10)^2]. Then st x =-3..y`=4.
3. y`=4 is the gradient :. b=4 in y=bx+3.
4. Find equation for tangent at (-3,-5.5,m=4) using y-y1=m(x-x1). Becomes y=4x+6.5. As y=mx+c:….c=6.5.
Hence
A=-5.5, b=4, c=6.5 and tangent equation is y=4x+6.5

Originally Posted by Neverquit

I'm looking for other coordinates on the curve where the tangent is parallel to y=bx+3. I have solved for a, b, c and tangent equation.

Question.
(-3,a) is on curve y=5x-7/2x+10.


Let us assume you mean . Not a good idea to be cheap with parentheses...

Thus,



The tangent at (-3,a) is parallel to y=bx+3 and cuts the y-axis at (0,c).

The tangent at has a slope equal to , and

The number we've been getting are nasty so perhaps that is not the function you meant. Check this.

Tonio[/color]



How do I get the other coordinates on the curve where the tangent is parallel to y=bx+3?
I’m sure I need to use the gradient perpendicular to the tangent (m=-1/4) and set it equal to y=5x-7/2x+10 or y`equation and solve for x?
Any help would be appreciated.


Workings
I have solved a,b,c (-5.5,4,-6.5) and the tangent equation y=4x-6.5. But graphing the curve shows me there is 2 hyperbolas so I need the points on the curve for the other tangent equation.

1. st x = -3 into y=(5x-7)/(2x+10) to get -5.5=a ;. (-3,-5.5).
2. find y` of y=(5x-7)/(2x+10). Use quotient rule..vu`-uv`/v^2.
Then you get [5(2x+10)-2(5x-7)]/[(2x+10)^2]. Then st x =-3..y`=4.
3. y`=4 is the gradient :. b=4 in y=bx+3.
4. Find equation for tangent at (-3,-5.5,m=4) using y-y1=m(x-x1). Becomes y=4x-6.5. As y=mx+c:….c=-6.5.
Hence
A=-5.5, b=4, c=-6.5 and tangent equation is y=4x-6.5

.

y=(5x-7)/(2x+10) is the correct function

not

y=4x-6.5. As y=mx+c:….c=-6.5. [/size][/font]
Hence
A=-5.5, b=4, c=-6.5 and tangent equation is y=4x-6.5

should be

Becomes y=4x+6.5. As y=mx+c:….c=6.5. [/size][/font]
Hence
A=-5.5, b=4, c=6.5 and tangent equation is y=4x+6.5

Originally Posted by Neverquit

y=(5x-7)/(2x+10) is the correct function

not

Yeah, too bad. A lesson to be learned for next time: either you type in LaTeX or else you write with enough parentheses to make completely clear your intention.

Anyway, the way's been shown, just do the corresponding corrections with the correct function.

Tonio

The coordinates of the other point on the curve where the tangent is parallel to y=bx+3. The answer is (-7,10.5).

I'm still unsure how to get -7. I know that if I s.t. y=10.5 into y=(5x-7)/(2x+10) I get -7.

How do I get the -7?