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 $\displaystyle y=x^2$ passing through $\displaystyle (1,-4)$.
$\displaystyle y=x^2$
$\displaystyle \Rightarrow \frac{dy}{dx}=2x$ ...(i)
(1)Let the tangent pass through the point $\displaystyle (a,b)$ such that $\displaystyle (a,b)$ lies on $\displaystyle y=x^2$, then the slope of the tangent is $\displaystyle m=2a$(using (i)). Since $\displaystyle (1,-4)$ lies on the tangent, the equation of the tangent is $\displaystyle y=2ax-a^2$ or $\displaystyle y=2ax-(2a+4)$.
(2)Also note that slope, $\displaystyle m=\frac{b+4}{a-1}=\frac{a^2+4}{a-1}$. We also know that $\displaystyle m=2a$ (we found it in point 1 using differentiation).
Relating these two: $\displaystyle 2a=\frac{a^2+4}{a-1}$
$\displaystyle \Rightarrow a^2-2a-4=0$.
Solving this with quadratic formula :$\displaystyle a=1 \pm \sqrt{5}$. Since there are two values of $\displaystyle a$, there must be two equations.
(3)Put the values of a in any of the two equations (in point 1) and you will get: $\displaystyle y=(2+2\sqrt{5})x-(6+2\sqrt{5})$ and $\displaystyle y=(2-2\sqrt{5})x+(-6+2\sqrt{5})$.
As we know, the tangent equation is the straight line $\displaystyle y=ax+b$.
From this "...through the point (1,-4)" we get the condition $\displaystyle a+b=-4$ (why?).
Also, the equations $\displaystyle y=x^2,~y=ax+b$ must have one common point; therefore, the quadratic equation $\displaystyle x^2=ax+b$ must have a single root, which is possible, if his discriminant is equal to zero $\displaystyle \Delta=a^2+4b=0$.
So, the unknown parameters of the tangent equations is solutions of this equations system
$\displaystyle \begin{cases}a+b=-4,\\ a^2+4b=0.\end{cases}$ which has two pairs of roots $\displaystyle \begin{cases}a_{1,2}=2\pm2\sqrt{5},\\ b_{1,2}=-6\mp2\sqrt{5}.\end{cases}$
Finally, we have $\displaystyle y=(2+2\sqrt{5})x+(-6-2\sqrt{5})$ and $\displaystyle y=(2-2\sqrt{5})x+(-6+2\sqrt{5})$.
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: $\displaystyle y= x^2= m(x- 1)- 4$ which gives the quadratic equation $\displaystyle y^2- mx+ (m+ 4)= 0$.
But in order to be tangent, that x must be a double root of the equation. And that means the "discriminant", $\displaystyle b^2- 4ac= (m^2)- 4(1)(m+4)= 0$. That gives a quadratic equation for m which has two roots as sbhatnagar and deMath indicate.