Drawing two perpendicular tangent line from the origin to $y=x^2-2x+a$

Question

We drew two perpendicular tangent line from the origin (the point $(0,0)$ ) to the curve $y=x^2-2x+a$. what is the value of $a$?

$1)\frac54\qquad\qquad2)\frac{-5}4\qquad\qquad3)\frac34\qquad\qquad4)\frac{-3}4$

At first I realized that we can write the quadratic equation as $y=(x-1)^2+(a-1)$ so the graph is obtained by moving $y=(x-1)^2$ upward or downward by $a-1$.

Also at the point $x=\alpha$ on the curve, the slope of the tangent line is $2\alpha-2$ . And I also know that the multiply of the slopes of perpendicular tangent lines is $-1$.

But I'm not sure how should I use these information to solve the problem.

Answer 1

You don't need to calculate the tangent lines concretely, you can do it by the process of elimination:

The three values $-\frac{5}{4}, -\frac{3}{4}$ and $\frac{3}{4}$ are not possible. Why? The peak is then below $0$. That means that one tangent has to point below the $x$-axis, however, the other tangent line has to reach the left side of the parabola and therefore go beyond the $y$-axis to the left. That means no angle of $90$ degree is possible.

In fact, for the first two values, you cannot even draw any tangent from $(0,0)$ as this point will lie above the parabola.

That means the only possibility is $a = \frac{5}{4}$. I've added an image of GeoGebra to show that this really works out

Answer 2

Let the tangent be $y=mx$, (as it passes through the origin), then we have the quadratic $x^2-(m+2)x+4-4a=0.$ In order for the given line be tangent to the given parabola, it must intersect at only one point, and hence ($B^2-4AC=0).$ So we get a quadratic for $m$ as $m^2+2m-4(1-a)=0.$ This means that two tangents are possible, and they will be perpendicular if $m_1m_2=-1 \implies 4(1-a)=-1 \implies a=5/4$ (product of roots of a quadratic $ax^2+bx+c = \frac{c}{a}$)

Answer 3

Let $f(x)=x^2-2x+a\require{cancel}$.

A line passing through $(0,0)$ is of the form $y=\lambda x$. If such a line is tangent to the graph of $f$ at some point $(\alpha,\beta)$, then:

1. $\beta=f(\alpha)=\alpha^2-2\alpha+a$;
2. $\lambda=f'(\alpha)=2\alpha-2$.

So, you have$$\beta=\lambda\alpha=2\alpha^2-2\alpha\quad\text{and}\quad\beta=\alpha^2-2\alpha+a,$$and therefore $2\alpha^2\cancel{-2\alpha}=\alpha^2\cancel{-2\alpha}+a$. So, $\alpha=\sqrt a$ or $\alpha=-\sqrt a$. In the first case, $\beta=2a-2\sqrt a$, and in the second case $\beta=2a+2\sqrt a$. In the first case, the slope of the tangent line will be $2\sqrt a-2$, and in the second case it will be $-2\sqrt a-2$. The only situation in which the tangent lines will be orthogonal, that is, the only case in which the product of these numbers is $-1$, is when $\sqrt a=\pm\sqrt{\frac54}$. Therefore, $a=\frac54$.

Answer 4

You had a good part of what you needed when you completed the square for the quadratic polynomial, in which you established that the vertex of the corresponding parabola has the coordinates $ \ (1 \ , \ a - 1) \ \ . $ So the axis of symmetry of this "upward-opening" parabola is "to the right" of the $ \ y-$axis, which requires that the tangent line from the origin to the parabola "to the left" of that axis be negative (the graph LegNaiB provides illustrates this). The slope of the other tangent line from the origin must therefore be positive; as the vertex is on or above this tangent line, we conclude that $ \ (a - 1) \ $ is positive, which is only true for choice $ \mathbf{(1)} \ \ . $

We can cinch this by considering the tangent points on the parabola. You found that $ \ \frac{dy}{dx} \ = \ 2x - 2 \ \ , $ so the tangent point "to the right" of the $ \ y-$ axis, $ \ x_1 > 0 \ , $ is given by $$ x^2_1 - 2x_1 + a \ \ = \ \ (2x_1 - 2) · x_1 \ \ \Rightarrow \ \ a \ = \ x^2_1 \ \ \Rightarrow \ \ x_1 \ \ = \ \ \sqrt{a} \ \ . $$ This tangent line then has the slope $ \ m \ = \ 2·(\sqrt{a} - 1) \ > \ 0 \ \ , $ which makes it plain that $ \ a > 1 \ \ . $

[Were this not a multiple-choice problem, we might go on to determine the second tangent point, $ \ x_2 < 0 \ , $ by finding intersection(s) of the parabola with the tangent line of negative slope, thus: $$ x^2_2 - 2x_2 + a \ \ = \ \ -\frac{1}{2·(\sqrt{a} - 1)} · x_2 $$ $$ \Rightarrow \ \ 2·(\sqrt{a} - 1)·x^2_2 \ + \ (-4\sqrt{a} + 5)·x_2 \ + \ 2·a·(\sqrt{a} - 1) \ \ = \ \ 0 \ \ , $$ a quadratic equation for which the discriminant is $$ \Delta \ \ = \ \ (-4\sqrt{a} + 5)^2 \ - \ 16·a·(\sqrt{a} - 1)^2 \ \ = \ \ 32·a\sqrt{a} \ - \ 16a^2 \ - \ 40\sqrt{a} \ + \ 25 $$ $$ = \ \ (5 - 4a) \ · \ (4a - 8\sqrt{a} + 5) \ \ . $$ The discriminant has a rational zero, $ \ a \ = \ \frac54 \ \ , $ and the other two zeroes are complex numbers. There is then a single intersection point of the line from the origin for this value of $ \ a \ \ , $ again confirming choice $ \mathbf{(1)} \ \ . $ The tangent lines are therefore $ \ y \ = \ (\sqrt5 - 2)·x \ $ and $ \ y \ = \ -(\sqrt5 + 2)·x \ \ , $ with the tangent points at $ \ x_1 \ = \ \frac{\sqrt5}{2} \ \ $ (thus, "to the right" of the vertex) and $ \ x_2 \ = \ -\frac{\sqrt5}{2} \ \ . $ Curiously, the tangent points are at equal "horizontal distances" from the $ \ y-$axis, despite the asymmetrical location of the parabola.]