derivatives
Find the coordinates of the point where the tangent to the curve $y=x^2+1$ at the point $(2,5)$ meets the normal to the same curve at the point $(1,2).$
I tried to form 2 equations for each set of coordinates given, then solve then simultaneously to get the 'y and x' coordinates needed, however I didn't seem to get the answer right.
Pick some point $(a,b)$ which lies on the curve $y=\frac{x}{x+1}$. This means that $b=\frac{a}{a+1}$. Taking the derivative using the quotient rule, we find:
$$y'(x)=\frac{1}{(1+x)^2}$$
so the slope of the tangent line at our point $(a,b)$ is
$$y'(a)=\frac{1}{(1+a)^2}$$
The point slope formula then tells us that the equation of the tangent line to the curve at the point $(a,b)$ is
$$y-b=\frac{1}{(1+a)^2}(x-a)$$
We want the point $(1,2)$ to satisfy this equation, so remembering that $b=\frac{a}{a+1}$, we have:
$$2-b=2-\frac{a}{a+1}=\frac{1}{(1+a)^2}(1-a)$$
This equation reduces to the quadratic $a^2+4a+1=0$ (which is the quadratic you found). The solutions $a=-2\pm\sqrt{3}$ can be plugged back into the equation
$$y=\frac{1}{(1+a)^2}(x-a)+\frac{a}{1+a}=\frac{1}{(1+a)^2}x+\frac{a^2}{(1+a)^2}$$
to obtain the equations of the desired tangent lines.
Here is a helpful picture plotted with Wolfram|Alpha:
Notice where the tangent lines intersect!
Since the curve is defined by the equation $$ F(x,y)=x^2+3xy-4y^2=0, $$ the line that is normal to it at the point $p=(6,6)$ is given by $$ N_p=\{(6,6)+t\nabla F(6,6): t \in \mathbb{R}\}=\{(6+t,6-t):\ t \in \mathbb{R}\}. $$ The problem is to find some $t \ne 0$ such that $$ F(6+t,6-t)=0. $$ Since \begin{eqnarray} F(6+t,6-t)&=&(6+t)^2+3(6+t)(6-t)-4(6-t)^2\\ &=&t^2+12t+36+3(36-t^2)-4(t^2-12t+36)\\ &=&-6t^2+60t=-60t(t-10), \end{eqnarray} we have $$ F(6+t,6-t)=0, t\ne 0 \iff t=10, $$ and the corresponding point of intersection is $p'=(16,-4)$.
Best Answer
$(y - y_0) = m_1(x - x_0)\tag{Point-Slope Equation Of Line}$
To find the slope of the normal line at the point $(1, 2)$, first evaluate $f'(1) = m_2= 2$, the slope of the tangent line at that point. Then the slope of the normal line to the curve at that same point is $m_\perp = -\dfrac 1{m_2} = -\dfrac 12$. Using the slope of the normal line $m_\perp = -\dfrac 12$ and the point $(x_0, y_0) = (1, 2)$, use the slope point form of an equation to obtain the equation of the line normal to the curve at $(1, 2)$:
$$(y - y_0) = m_\perp(x - x_0)$$
Finally, find where the tangent line and the normal line intersect by setting the equations equal to one another. Solve for $x$, then find the corresponding $y$ value using the equation of either line.