Typo in book or I am wrong?

Question

Find the equation of the tangent to the parabola $y = x^2$, if the $x$-intercept of the tangent is $2$.

Now $$y = mx + b$$ $$0 = m(2) + 2$$ $$m = -1$$

so $$m = \frac{y-0}{x-2}$$ $$m(x-2) = y$$ $$-x-y = -2$$

but the book answer is

$$8x-y =16$$

Answer 1

Say $(X,Y)$ is the point where the line is tangent to the parabola. Then $Y=X^2$. Furthermore, using calculus (one of the question tags), the tangent line has slope $2X$, and we are given that the $x$-intercept is $2$.

So the equation for the tangent line is $y=2Xx+b$, which becomes $X^2=2X^2+b$ at the tangent point and $0=2X\cdot2+b$ at the $x$-intercept, so $0=X^2+b$ and $0=4X+b$. From there we see that $X=4$ (or $0$, but then the tangent line would not have a unique $x$-intercept) and $b=-16$, and so the equation for the tangent line is $y=8x-16$.

Answer 2

$y = mx + b$

$(2,0)$ lies on the line.

$0 = 2m + b\\ b = -2m$

Here is the error in the original post.

$y = mx - 2m$ is tangent to $y = x^2$

$x^2 = mx - 2m\\ x^2 - mx + 2m = 0$

We must find the value of $m$ such that there is exactly one solution to this quadratic.

We can use the quadratic formula:

$x = \frac {-m \pm \sqrt {m^2 - 8m}}{2}$

In order to have 1 solution $m^2 - 8m = 0$

$m = 0$ or $m = 8$

Case 1: $m=0$
$y= 0$ is tangent to the curve and does go through the point $(2,0)$ but as it represents the entire x-axis, it would not be accurate to say that this line has an x-intercept.

Case 2: $m = 8$
$y = 8x - 16$ which is equivalent to the book's solution.

Answer 3

In general for $y=f(x)$ the line tangent at $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a).$ For $f(x)=x^2$ we get $y-a^2=2a(x-a).$ The slope must be nonzero, i.e. $a\neq 0.$ When $y=0$ then $x={a\over 2}.$ Hence $a=4$ and the equation of the tangent line is $y-16=8(x-4),$ i.e. $y=8x-16.$

Answer 4

Book answer is right.

We want $m$, $b$, $x_0$ such that $y=mx+b$ touches the parabola $y=x^2$ at point $x_0$.

We know $\frac{\partial}{\partial x}x^2=2x$, so since the slope will be tangent to the parabola at the kissing point we know $m=2x_0$. From $0=2m+b$ we can conclude $b=-4x_0$. Now we find $x_0$: \begin{align} \frac{y_0-y_\text{intercept}}{x_0-x_\text{intercept}}&=m\\ \frac{y_0-0}{x_0-2}&=2x_0\\ y_0&=2x_0^2-4x_0\\ x_0^2&=2x_0^2-4x_0\\ 0&=x_0^2-4x_0\\ x_0&=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(0)}}{2(1)}\\ &=\frac{4\pm\sqrt{16+0}}{2}\\ &=4,0 \end{align} We can discard $x_0=0$ as a viable solution. Now we have: \begin{align} m&=2x_0=8\\ b&=-4x_0=-16\\ y&=mx+b\\ y&=8x-16\\ 8x-y&=16 \end{align}