It is given that the tangent to curve at points $x=1$ and $x=-1$ are perpendicular. I've managed to find the equation of the curve: y=$\frac{4}{3}x- \frac{5}{6}x^2$ but how do I show that the curve passes through the origin?
[Math] How to show that a curve passes through the origin
algebra-precalculus
Related Solutions
"Passing through" means that it is on the circle i.e. it is one of the points on the circle.
How to solve:
You are given a point on the circle, $(-2, 1)$, a tangent line $3x - 2y = 6$ and a tangent point $(4, 3)$.
Find the slope of the tangent line:
$3x - 2y = 6$
$-2y = -3x + 6$
$y = \frac{3}{2}x + 3$
The slope is $\frac{3}{2}$.
Recall that the slope of a perpendicular line is the negative reciprocal of the original line. In other words, if $m_1$ is the slope of a line, and $m_2$ is the slope of the perpendicular line, then $m_1 * m_2 = -1$.
The radius must be perpendicular to the tangent line. Using the above formula, let $m_1$ be equal to the slope of the tangent line, $\frac{3}{2}$. Then $m_2$, the slope of the radius, can be found like so:
$\frac{3}{2} * m_2 = -1$
$m_2 = -\cfrac{1}{\left(\cfrac{3}{2}\right)}$
$m_2 = -1 * \frac{2}{3}$
$m_2 = -\frac{2}{3}$
Using point-slope form, derive an equation of a line that the centre is on.
$y - y_0 = m(x - x_0)$
$y - 3 = -\frac{2}{3}(x - 4)$
$y = -\frac{2}{3}x - \frac{8}{3} + 3$
$y = -\frac{2}{3} + \frac{1}{3}$
Let the coordinates of the center be $(a, b)$. Then the equation of the circle is $(x-a)^2 + (y-b)^2 = r^2$
Remember that $b = -\frac{2}{3}a + \frac{1}{3}$
Substitute $b$ for the expression above, and plug in the point $(-2, 1)$ into the equation.
$(-2-a)^2 + (1-(-\frac{2}{3}a + \frac{1}{3}))^2 = r^2$
$a^2 + 4a + 4 + (\frac{2}{3}a - \frac{2}{3})^2 = r^2$
$a^2 + 4a + 4 + \frac{4}{9}a^2 - \frac{8}{9}a + \frac{4}{9} = r^2$
$\frac{13}{9}a^2 + \frac{28}{9}a + \frac{40}{9} = r^2$
Do the same thing, but instead, plug in the point $(4, 3)$ into the equation.
To make this faster, $(4-a)^2 + (3-(\frac{2}{3}a + \frac{1}{3}))^2$ evaluates to $a^2 - 8a + 16 + (-\frac{2}{3}a + \frac{8}{3})^2$, which equals $a^2 - 8a + 16 + \frac{4}{9}a^2 - \frac{32}{9}a + \frac{64}{9}$. Further simplification yields $\frac{13}{9}a^2 - \frac{40}{9}a + \frac{80}{9}$.
Now you have two equations in two variables. This creates a system.
$\left\{ \begin{array}{rcl} \frac{13}{9}a^2 + \frac{28}{9}a + \frac{40}{9} = r^2 \\ \frac{13}{9}a^2 - \frac{40}{9}a + \frac{80}{9} = r^2 \\ \end{array} \right.$
Multiply both equations by $9$ to get rid of the fractions. This allows for easier work.
$\left\{ \begin{array}{align} 13a^2 + 28a + 40 = 9r^2 \\ 13a^2 - 9a + 80 = 9r^2 \\ \end{array} \right.$
Equate the expressions and solve.
$13a^2 + 28a + 40 = 13a^2 - 9a + 80$
$37a - 40 = 0$
$37a = 40$
$a = \frac{40}{37}$
Plug in the value of $a$ into the equation $b = -\frac{2}{3}a + \frac{1}{3}$ defined above.
$b = -\frac{2}{3} * \frac{40}{37} + \frac{1}{3}$
$b = -\frac{80}{111} + \frac{1}{3}$
$b = -\frac{80}{111} + \frac{37}{111}$
$b = -\frac{57}{111}$
The center of the circle is at $(\frac{40}{37}, -\frac{57}{111})$.
To find $r^2$, use the distance formula, but square both sides so that the square root is removed. Allow me to demonstrate:
Original distance formula: $D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$, where $D$ is the radius $r$, and $(x_1, y_1)$ and $(x_2, y_2)$ are any two given points.
Squaring both sides gives $(x_1 - x_2)^2 + (y_1 - y_2)^2 = r^2$, which looks a lot like the equation of a circle (it actually is not).
Plug in the points $(\frac{40}{37}, -\frac{57}{111})$ and $(-2, 1)$ into the formula.
$(\frac{40}{37} + 2)^2 + (-\frac{57}{111} - 1)^2 = r^2$
$(\frac{114}{37})^2 + (-\frac{168}{111})^2 = r^2$
$\frac{12996}{1369} + \frac{28224}{12321} = r^2$
$\frac{198762372}{16867449} = r^2$
The equation of the circle is:
$(x - \frac{40}{37})^2 + (y + \frac{57}{111})^2 = \frac{198762372}{16867449}$
I might have made a calculation mistake / error, but you get the idea of how to do it. I hope this answer helped you!
EDIT: Yes I did make a calculation error when I was solving the system, but you get the point. DO NOT use my answer for your assignment / homework, because it is wrong.
Notice, in general, the equation of the line passing through the point $(x_1, y_1)$ & having slope $m$ is given by the point-slope form: $$\color{blue}{y-y_1=m(x-x_!)}$$
We have, equation of line A: $3x+6y-1=0\iff \color{blue}{y=-\frac{1}{2}+\frac{1}{6}}$ having slope $-\frac{1}{2}$
1.) slope of the line passing through $(5, 1)$ & perpendicular to the line A $$=\frac{-1}{\text{slope of line A}}=\frac{-1}{-\frac{1}{2}}=2$$ Hence, the equation of the line: $$y-1=2(x-5)$$ $$y-1=2x-10$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{2x-y-9=0}}$$
2.) slope of the line passing through $(5, 1)$ & parallel to the line A $$=\text{slope of line A}=-\frac{1}{2}$$ Hence, the equation of the line: $$y-1=-\frac{1}{2}(x-5)$$ $$2y-2=-x+5$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x+2y-7=0}}$$
Best Answer
I'm assuming the equation is $y=\tfrac 43 x- \tfrac 56 x^2$.
A curve passes through the origin if the point $(0,0)$ satisfies the equation:
$$0 \overset{?}{=} \tfrac 43 (0)- \tfrac 56 (0)^2$$