Chapter 11, Problem 5P

Tangent Lines to a Parabola In this problem we show that the line tangent to the parabola y = x² at the point (a, a²) has the equation y = 2ax − a².

1. (a) Let m be the slope of the tangent line at (a, a²). Show that the equation of the tangent line is y − a² = m(x − a).
2. (b) Use the fact that the tangent line intersects the parabola at only one point to show that (a, a²) is the only solution of the system.

$\{\begin{array}{l}y-a^2=m(x-a)\hfill \\ y=x^2\hfill \end{array}$

1. (c) Eliminate y from the system in part (b) to get a quadratic equation in x. Show that the discriminant of this quadratic is (m − 2a)². Since the system in part (b) has exactly one solution, the discriminant must equal 0. Find m.
2. (d) Substitute the value for m you found in part (c) into the equation in part (a), and simplify to get the equation of the tangent line.

To determine

To show: The equation of the tangent line to the parabola $y=x^2$ at the point $\left(a,a^2\right)$ is $y-a^2=m\left(x-a\right)$.

Explanation of Solution

Formula used:

“Let $y=f\left(x\right)$ be a curve and $p\left(x_1,y_1\right)$ be a point on the curve. Then,

Slope of the tangent at p is $m={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}$.

The equation of tangent at p is $y-y_1=m\left(x-x_1\right)$”.

Given the equation of the parabola is $y=x^2$.

Let $m$ be the slope of the tangent at $\left(a,a^2\right)$.

Use the above mentioned formula and obtain the equation of the tangent at $\left(a,a^2\right)$.

Substitute $x_1=a$ and $y_1=a^2$ in $y-y_1=m\left(x-x_1\right)$,

$y-a^2=m\left(x-a\right)$

Thus, it is shown that the equation of tangent line is $y-a^2=m\left(x-a\right)$.

To determine

To show: The point $\left(a,a^2\right)$ is the only solution of the system $\{\begin{array}{l}y-a^2=m\left(x-a\right)\\ y=x^2\end{array}$.

Explanation of Solution

Given the system of equations are $y-a^2=m\left(x-a\right)$ and $y=x^2$.

If $y-a^2=m\left(x-a\right)$ is a tangent to the $y=x^2$, then the tangent line touches the $y=x^2$ at only one point. That is, $y-a^2=m\left(x-a\right)$ and $y=x^2$ have only one solution.

Solve $y-a^2=m\left(x-a\right)$ and $y=x^2$ to get point of intersection $p\left(x_1,y_1\right)$. At the point $p$, the equations become $y_1=x_1{}^2$ and $y_1-a^2=m\left(x_1-a\right)$.

Substitute $y_1=x_1{}^2$ in $y_1-a^2=m\left(x_1-a\right)$,

$\begin{array}{c}\left(x_1{}^2-a^2\right)=m\left(x_1-a\right)\\ \left(x_1+a\right)\left(x_1-a\right)=m\left(x_1-a\right)\\ x_1+a=m\\ x_1=m-a\end{array}$

Now, differentiate $y=x^2$ with respect to x,

$\frac{dy}{dx}=2x$

Slope of the tangent at $p\left(x_1,y_1\right)$,

$\begin{array}{l}m={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\\ m=2x_1\end{array}$

Now substitute $m=2x_1$ in $x_1=m-a$,

$\begin{array}{c}{x}_1=2x_1-a\\ -x_1=-a\\ x_1=a\end{array}$

Substitute $x_1=a$ in the equation $y_1=x_1{}^2$,

$y_1=a^2$

Thus, the point $\left(a,a^2\right)$ is the only solution.

Thus, it is shown that $\left(a,a^2\right)$ is the only solution of the system, $\{\begin{array}{l}y-a^2=m\left(x-a\right)\\ y=x^2\end{array}$.

To determine

To show: The discriminant of the quadratic equation is ${\left(m-2a\right)}^2$ and find the value of m.

Answer to Problem 5P

The value of m is $2a$.

Explanation of Solution

Formula used:

(1) The discriminant of the quadratic equation $A{x}^2+Bx+C=0$ is $B^2-4AC$.

(2) If the quadratic equation $A{x}^2+Bx+C=0$ has only one solution, then the discriminant is zero.

Calculation:

Given the system of equations are $y-a^2=m\left(x-a\right)$ and $y=x^2$.

Substitute $y=x^2$ in $y-a^2=m\left(x-a\right)$,

$\begin{array}{c}{x}^2-a^2=m\left(x-a\right)\\ \left(x^2-a^2\right)=mx-am\\ x^2-a^2-mx+am=0\end{array}$

The above equation does not contain the variable $y$, thus $y$ is eliminated from the system $\{\begin{array}{l}y-a^2=m\left(x-a\right)\\ y=x^2\end{array}$.

Thus, the resulting quadratic equation is $x^2-a^2-mx+am=0$.

Compare the quadratic equation $x^2-a^2-mx+am=0$ with $A{x}^2+Bx+C=0$,

It is clear that $A=1$, $B=-m$, $C=am-a^2$.

Use the discriminant formula to compute the value of the discriminant.

$\begin{array}{c}{B}^2-4AC={\left(-m\right)}^2-4\left(1\right)\left(am-a^2\right)\\ =m^2-4am+4a^2\\ =m^2-2\left(m\right)\left(2a\right)+{\left(2a\right)}^2\\ ={\left(m-2a\right)}^2\end{array}$

Thus, it is shown that the discriminant of the quadratic equation $x^2-a^2-mx+am=0$ is ${\left(m-2a\right)}^2$.

Since, the equation $y-a^2=m\left(x-a\right)$ and $y=x^2$ has exactly one solution then the discriminant $=0$ by the above mentioned formula.

$\begin{array}{c}{\left(m-2a\right)}^2=0\\ m-2a=0\\ m=2a\end{array}$

Thus, the value of $m$ is $2a$.

To determine

To find: The equation of the tangent line $y-a^2=m\left(x-a\right)$ by substituting the value of m.

Answer to Problem 5P

The equation of the tangent line is $y=2ax-a^2$.

Explanation of Solution

From part (c), the slope of the tangent is $m=2a$.

From part (a), the equation of the tangent is $y-a^2=m\left(x-a\right)$.

Substitute $m=2a$ in $y-a^2=m\left(x-a\right)$,

$\begin{array}{c}y-a^2=2a\left(x-a\right)\\ y-a^2=2ax-2a^2\\ y=2ax-a^2\end{array}$

Thus, the equation of tangent line is $y=2ax-a^2$.