Hello,
So I am having problems with trying to find the equations of tangent lines when the point the lines pass through is not on the original function.
ex:
f(x)=x^2
find equations of tangent lines through the point (1,-4)
Hello,
So I am having problems with trying to find the equations of tangent lines when the point the lines pass through is not on the original function.
ex:
f(x)=x^2
find equations of tangent lines through the point (1,-4)
we need to find the tangents to the curve passing through .
...(i)
(1)Let the tangent pass through the point such that lies on , then the slope of the tangent is (using (i)). Since lies on the tangent, the equation of the tangent is or .
(2)Also note that slope, . We also know that (we found it in point 1 using differentiation).
Relating these two:
.
Solving this with quadratic formula : . Since there are two values of , there must be two equations.
(3)Put the values of a in any of the two equations (in point 1) and you will get: and .
As we know, the tangent equation is the straight line .
From this "...through the point (1,-4)" we get the condition (why?).
Also, the equations must have one common point; therefore, the quadratic equation must have a single root, which is possible, if his discriminant is equal to zero .
So, the unknown parameters of the tangent equations is solutions of this equations system
which has two pairs of roots
Finally, we have and .
See the picture
As long as f is a polynomial, you can solve an problem like this using Fermat's method of "ad-equation" (predating Calculus). As has been said, the equation of any line through (1, -4) must be of the form y= m(x- 1)- 4. Also as has been said, at a point of tangency, the curve and line must cross: which gives the quadratic equation .
But in order to be tangent, that x must be a double root of the equation. And that means the "discriminant", . That gives a quadratic equation for m which has two roots as sbhatnagar and deMath indicate.