The line is tangent to a parabola

Question

The line $y = 4x-7$ is tangent to a parabola that has a $y$-intercept of $-3$ and the line $x=\frac{1}{2}$ as its axis of symmetry. Find the equation of the parabola. I really need help solving this question. THx

Answer 1

Hint: remember that the equation of a parabola is $y = ax^2 + bx +c$ and $c$ is the intersection with the axis y (that you know in your exercise). In addition, you have to solve a system between your generic parabola and the tangent line and impose that they are only one point of intersection (putting $\Delta = 0$). The last condition is $- \frac{b}{2a} =\frac{1}{2}$. In this way, you'll find $a$, $b$ and $c$.

It's simpler if you use before the condition $c = -3$ and $- \frac{b}{2a} =\frac{1}{2}$ and then solve the system.

Answer 2

Let the formula for the parabola be $f(x)=ax^2+bx+c$

1. y-intercept: $y$-intercept is @ $(0,-3)$ so we have: $f(0)=-3 \Rightarrow c=-3$.

2. Vertex: The vertex is @ $(x,\frac{1}{2})$ and we now that at the vertex, the derivative is equal to zero, so we have: $f'(\frac{1}{2})=0 \Leftrightarrow 2a\cdot(\frac{1}{2})+b=0 \Rightarrow a+b=0$.

3. Line tangent to parabola: Furthermore, the line $y=4x-7$ is tangent to the parabola (and therefore $4x-7=ax^2+bx+c$ has only one solution).

Intersection point: $ax^2+bx+c=4x-7 \Rightarrow ax^2+x(b-4)+(c+7)$. Discriminant must be zero in order to have only one intersection point, so we have $D=b^2-4ac=(b-4)^2-4a(c+7)=0$.

Now, we substitute $c=-3$ and $a=-b$ and solve for $a$:

$(b-4)^2+4b(-3+7)=0 \Rightarrow (b-4)^2+16b=0 \Rightarrow b=-4$. $a=-b$ so $a=4$.

The formula is $\boxed{y=4x^2-4x-3}$.

Answer 3

Let the parabola be defined by the function $y = f(x)$ and let $g(x) = f(x) - 4x + 7.$ Then $y = g(x)$ is a parabola passing through $(0,4)$ and it has slope $-4$ at $x = \frac 12.$ In addition, the parabola described by $y = g(x)$ is tangent to the $x$-axis so $g(x)$ can be written $g(x) = a(x - p)^2.$ We can also write the derivative of $g$ with respect to $x$ as $g^\prime(x) = 2a(x - p).$

This gives us $g(0) = ap^2 = 4$ and $g^\prime\left(\frac 12\right) = a(1 - 2p) = -4.$ Then $$ \frac{ap^2}{a(1 - 2p)} = \frac{4}{-4}, $$ and we can simplify both sides of this to conclude that $$ \frac{p^2}{1 - 2p} = -1. $$ Rearranging terms, $$ p^2 = -1(1 - 2p), $$ $$ p^2 - 2p + 1 = 0 $$ Solve for $p;$ that is the $x$-coordinate of the (unique) point where $g(x) = 0.$ It's also the $x$-coordinate of the unique point where $f(x) = 4x - 7.$ Given $p,$ use one of the previous equations (such as $ap^2 = 4$) to solve for $a.$ You can then expand $g(x)$ as a simple polynomial. Use the fact that $f(x) = g(x) + 4x - 7$ to write $f(x).$

You can solve the problem without introducing the additional parabola $y = g(x),$ but I like the geometric intuition of this method. The key idea is that by a change of coordinates ($y \rightarrow y - 4x + 7$), we transform the parabola $y = f(x)$ into one with a (somewhat) easier geometry (or at least so I think).

It seems a mere coincidence that the point of tangency of $y = f(x)$ and $y = 4x - 7$ just happens to be a point that we could already know is on $y = f(x)$ by virtue of the axis of symmetry and the one given point of that parabola. That's not a necessary condition for this problem to be solvable. It does suggest a clever insight that for this particular problem that leads to a solution more quickly than the steps above, because you can find the point of tangency almost right away.