Linear line that passes through the origin as a tangent to Quadratic line

I have a known quadratic line and I need a linear line that passes through the origin and is a tangent to my quadratic line. I originally tried to solve it by equating the quadratic to a linear line, ax^2 +bx+c=mx, rearranging to equal 0, 0= ax^2+(b-m)x+c. Solve for x, x=(-(b-m)+-sqrt((b-m)^2-4ac))/2a. For one solution the determinant (bit under the sqrt) must equal 0, 0=(b-m)^2-4ac, 0= b^2+m^2-2mb-4ac, solving for m should equal the gradient I need for a tangent, 0=m^2-2mb+(b^2-4ac). However this gives me an imaginary number and I can't understand it. Code and line used below

y=-0.008877*x^2+0.9859*x+5.509 % line, to speed up the process 
% I have written it out here, it is in a polyfit for some massive dataset

syms V0 m a b d
help=solve(b^2 - 2*b*m + m^2 - 4*a*d==0,m)
gradient=subs(help,[a,b,d],[-0.008877,0.9859,5.509])
eqn = eval(coeffs(gradient(2))) % I have tried both solutions and work it 
% out by hand equally an imaginary answer as well
tangentX=[-50:1:100]
tangentY=[-50:1:100].*(eqn)
hold on
plot(tangentX,tangentY,'k')

Answer 1

[Torsten]

The tangent to your quadratic through a point

(x0,a*x0^2+b*x0+c)

is given by

y(x) = (2*a*x0+b) * x + ((a*x0^2+b*x0+c) - (2*a*x0+b) * x0)

For the tangent to pass through (0,0), the y-intercept of the line must be equal to 0, thus

((a*x0^2+b*x0+c) - (2*a*x0+b) * x0) = 0.

Solve for x0.

[Katherine]

This does not provide a solution nor do I particularly understand why you've said waht you have said. Have you ran this code and did it work for my line?

expanding your last equation is ax^2+bx+c-2ax^2-2bx=0

-ax^2-bx+c=0

giving 2 real solutions of x=105, and x=-5.9013

neither give a tangent passing through 0

[Torsten]

What's the problem with my answer ?

Can't you calculate the equation of a tangent to f(x) = a*x^2+b*x+c in a point (x0,f(x0)) ?

Can't you follow that for the tangent to pass through (0,0), its y-intercept must equal 0 ?

For your given function there is no tangent passing through the origin, as you can see from the plot and the results for x0_tangent and y0_tangent which are complex-valued.

syms a b c x0
eqn = ((a*x0^2+b*x0+c) - (2*a*x0+b) * x0) == 0;
x_tangent = solve(eqn,x0)

x_tangent = 

y_tangent = a*x_tangent.^2+b*x_tangent+c

y_tangent = 

f=@(x)-0.008877*x.^2+0.9859*x+5.509;
x = -1000:0.01:1000;
plot(x,f(x))

format long
x0_tangent = double(subs(x_tangent,[a c],[-0.008877,5.509]))

x0_tangent = 
  0.000000000000000 -24.911694894680050i
  0.000000000000000 +24.911694894680050i

y0_tangent = double(subs(y_tangent,[a b c],[-0.008877 0.9859 5.509]))

y0_tangent = 
 11.018000000000001 -24.560439996665060i
 11.018000000000001 +24.560439996665060i

[Katherine]

Your answer is poorly explained. Can't you understand that? Can't you try to run your own code and realise the answer is still imaginary? and therefore not the answer I am looking for and if the only solution is imaginary, please can you explain why?

As highlighted in the original question, I quite clearly know the intercept of my linear line will be 0. I literally walk through your steps in my original solution. Can't you read it and understand I solve it very similarly but get an imaginary answer?

[Katherine]

Anyway someone has explained this to me, that my line has the peak above the origin and therefore if I want my tangent line to pass through 0, it in "inside" the curve and therefore I keep getting imagiary numbers because there is no solution as the line, that is a tangent and passes through the origin can't exist. Therefore there isn't a problem with my maths or my understanding. It is a problem with the original quadratic and since it is based on experimental data, my environmental factors, human error etc. msut be too large to produce good data and the experiment must be rerun

[Torsten]

You have the quadratic

f(x) = a*x^2 + b*x + c

Now I want to calculate the slope in a point (x0,f(x0));

f'(x) = 2*a*x + b

evaluated in x0:

slope = 2*a*x0 + b.

Now the equation of the tangent in (x0,f(x0)) is

t(x) = slope*x + n

To determine n, you know that t passes through (x0, a*x0^2+b*x0+c), thus

a*x0^2+b*x0+c = t(x0) = slope*x0 + n

thus

n = -slope*x0 + (a*x0^2+b*x0+c).

Thus the equation of the tangent through (x0,a*x0^2+b*x0+c) is

t(x) = slope*x + (-slope*x0 + (a*x0^2+b*x0+c))

Now you want to determine a point x0 for which the tangent passes through (0,0).

For this, the y-intercept of the tangent t

-slope*x0 + (a*x0^2+b*x0+c) = -(2*a*x0 + b)*x0 + (a*x0^2+b*x0+c)

must be 0.

Now you can continue with my solution with the symbolic toolbox.

The fact that x0_tangent and y0_tangent become complex shows that there does not exist a real-valued point (x0,f(x0)) on your quadratic such that the tangent in this point passes through the origin. This can easily be seen from the graph of your quadratic.

Most probably, this will be the reason with your code, too, but I could not follow your deduction of computing m.

[Katherine]

M is the gradient of the linear line, a standard linear line equation is y=mx+c, so I just used m as the symbol for the gradient, since I know the y intercept is 0 c=0. I know there is 1 intercept between the quadratic and the linear so I equated, them and used the quadratic formula to solve for x, since I also know there is a singular solution for x as the line is a tangent, this means the determinant of the quadratic formula =0, the determinant of the quadratic formula in this case is a quadratic involving m, solving this will give the gradient of the linear line. In this case it gives an imaginary number and I know why now. I thought originally my code might be wrong but I am fairly confident as it works for multiple other lines, so I came to the resolution it must be with the data which considering it worked for other experiments on the same day, so little change in the conditions, I found this hard to believe but must be the only explanation as to why the code doesn't work because math doesn't lie but experiments do XD

Thank you for taking the time to fully explain your method.