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The equation of the circle is given by $$\left(x-\frac{x_1+x_2}{2}\right)^2+\left(y-\frac{y_1+y_2}{2}\right)^2=\frac 14\left((x_1-x_2)^2+(y_1-y_2)^2\right)$$ From what you've got, you can write it as $$\left(x-\left(a+\frac{2a}{m^2}\right)\right)^2+\left(y-\frac{2a}{m}\right)^2=4a^2\left(\frac{1}{m^2}+1\right)^2$$ Now, set $x=-a$ which is the equation of the ...


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HINT: Using this OR this, WLOG the endpoints of one diameter the relevant circle can be written as $$(at^2,2at), (a/t^2,-2a/t)$$ Use this, to form the equation of the circle Now, as the equation of directrix is $x+a=0,$ put $x=-a$ in the equation of the circle


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Justification of the formula: After centering (translation to let the linear terms vanish), the equation becomes $$Ax^2+Bxy+Cy^2+F'=0.$$ Then you apply a rotation of angle $\theta$ to let the mixed term $Bxy$ vanish from the quadratic terms, $$A(x\cos\theta-y\sin\theta)^2+B(x\cos\theta-y\sin\theta)(x\sin\theta+y\cos\theta)+C(x\sin\theta+y\cos\theta)^2.$$ ...


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There're two principal axes in general, so \begin{align*} \theta &=\frac{1}{2} \tan^{-1} \frac{B}{A-C}+\frac{n\pi}{2} \\ &= \tan^{-1} \left( \frac{C-A}{B} \color{red}{\pm} \frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \: \right) \\ \end{align*} The centre is given by $$(h,k)= \left( \frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC} \right)$$ ...


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Although it sounds like a question, for calculation did you use atan2 function or atan function? Quadrant placement is also important.


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Hint: By symmetry, a quadric with two focal points is an ellipsoid of revolution and can't be more. In 2D, a conic is defined by $5$ independent parameters, which can be the coordinates of two foci and the sum of distances. A general quadric needs $9$ parameters, which cannot equal $3n+1$. (Not a proof but a serious clue; similarly, the reduced conic has ...


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For confocal central quadrics, $$\frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}+\frac{z^2}{c^2+s}=1$$ where $a>b>c\geq 0$, there're one focal ellipse namely, $$\left \{ \begin{array}{rcl} \displaystyle \frac{x^2}{a^2-c^2}+\frac{y^2}{b^2-c^2} &=& 1 \\ z &=& 0 \end{array} \right.$$ and one focal hyperbola namely, $$\left \{ \begin{array}{...


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The answer depends on the number of prime factors of $k$ when we consider a factorization over the ring of Eisenstein integers $\mathbb{Z}[\omega]$. Have a look at this answer, too. For instance, there are just trivial solutions if $k$ is a square-free number and a product of primes of the form $3m-1$, that do not split over $\mathbb{Z}[\omega]$. So the ...


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I found the answer. Here goes.. By letting $A$ be the point of reference we have: $\vec{OM}(\vec{OM}-2\vec{OA})=7 \implies (\vec{AM}-\vec{AO})(\vec{AM}-\vec{AO}-2\vec{OA})=7 \implies (\vec{AM}-\vec{AO})(\vec{AM}+\vec{AO})=7 \implies |\vec{AM}|^2 -|\vec{AO}|^2=7 \implies |\vec{AM}|^2=7+9=16 \implies |\vec{AM}|=4$ Therefore we notice that the points $M$ are ...


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There are several definitions of ellipticity (also called eccentricity), though they aren't different things quite so much as different measurements of divergence from perfect circularity: Your first definition is the first (or "primary") flattening f, which is equal to the versine of the angle [ref. 1]; Your second definition is the first (or "primary") ...


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Proof of Focus Focus: Sphere $G_1$ is tangent to the plane the ellipse is contained on at point $F_1$. It is also tangent to the cone at circle $k_1$ Sphere $G_2$ is tangent to the same plane at point $F_2$. It is also tangent to the cone at $k_2$. Connect the apex of the cone, $S$, to any point on the ellipse, $P$, with a line. Mark the intersection ...


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Let $(X,Y,Z)$ be the center of the circle $ABC$. This is on the plane $$\frac xa+\frac yb+\frac zc=1\tag1$$ so, we can write $$Z=C\left(1-\frac Xa-\frac Yb\right)$$ Solving $$(X-a)^2+Y^2+\left(C\left(1-\frac Xa-\frac Yb\right)\right)^2=X^2+(Y-b)^2+\left(C\left(1-\frac Xa-\frac Yb\right)\right)^2=X^2+Y^2+\left(C\left(1-\frac Xa-\frac Yb\right)-C\right)^2$$ ...


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The generator line of the first cone is $\frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}.$ Setting $y=0$ gives $$\frac{x-\alpha}{l_1}=\frac{0-\beta}{m_1}=\frac{z-\gamma}{n_1}\implies x=-\beta\frac{l_1}{m_1}+\alpha,\quad z=-\beta\frac{n_1}{m_1}+\gamma\tag1$$ Also, $$\frac{l_1}{m_1}=\frac{x-\alpha}{y-\beta},\quad \frac{n_1}{m_1}=\frac{z-\gamma}...


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[N.B.: This is only a partial solution for the reason given at the end.] If I understand the question correctly, given a rectangle aligned with the coordinate axes, you’d like to inscribe an ellipse in it that has a given inclination $\phi$. There are several constructions for ellipses inscribed in rectangles given at this web site. The Java applets there ...


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Now, to prove these definitions are equivalent, we’re going to take the first definition and make an equation out of it. We will then manipulate it a little. Then, we’ll do the same for the second definition and we’ll get the second equation into the form of the first equation. Once we’ve done that, we will prove that the eccentricities are equal for both ...


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There are two ways to prove this: Formal: You can show, through a bunch of ugly computation, that the expression $B^2-4AC$ is invariant under rotation. So, consider when $B=0$ (in other words, when the conic section's directrix is parallel to one of the axes). It is easy to see that for a hyperbola $-4AC$ is positive, for an ellipse $-4AC$ is negative, and ...


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As Robert Frost's hint suggests, there's a simple geometric solution. $$\frac{x}{a} + \frac{y}{b}= 1$$ is the intercept form of the equation of a line, i.e., the X & Y intercepts of the line are $a$ and $b$, respectively. Let $c$ be the length of the line segment between the intercept points. This segment is the hypotenuse of a right triangle, so $$c^...


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As achille hui wrote in a comment, your setup is invariant under affine transformations. So you can simplify things by choosing an affine coordinate system in such a way that the tangent lines meet at $(0,0)$ and the points of contact are $(1,0)$ and $(0,1)$. In that case, for reasons of symmetry the centers of all your ellipses have to lie on the line $x=y$....


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Here's a slightly different way. Rather than messing around with $x$ and $y$ we express the equation of the line in parametric form and work with the parameter $t$. The equation of the line is given in intercept form, i.e., the line passes through $(0, b)$ and $(a, 0)$. We can easily change it to parametric form: $$x = at, \, y = b(1 - t)$$ where $t = 0$ ...


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Using Vieta's formula should help. The distance between the line and the origin is $$\frac{|-ab|}{\sqrt{a^2+b^2}}$$ so, in the following, we may suppose that $$\frac{|-ab|}{\sqrt{a^2+b^2}}\le r.$$ Let $\alpha,\beta$ be the $x$ coordinates of the intersection points. Then, by Vieta's formula, $$\alpha+\beta=\frac{2ab^2}{a^2+b^2},\quad \alpha\beta=\frac{a^2(...


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Solving these two simultaneously gives me: $$x=\frac{a(b^2\pm \sqrt{r^2(a^2+b^2)-b^2})}{a^2+b^2}$$ $$y=\frac{b(a^2\pm \sqrt{r^2(a^2+b^2)-a^2})}{a^2+b^2}$$ To find the chord length we can find $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. This gives: $$\sqrt{\left(\frac{2a\sqrt{r^2(a^2+b^2)-b^2}}{a^2+b^2}\right)^2+\left(\frac{2b\sqrt{r^2(a^2+b^2)-a^2}}{a^2+b^2}\...


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The line of intersection of the planes through $OX$ and $OY$ does not lie on the $xy-$plane. Let $n_1$ and $n_2$ be the unit vectors perpendicular to the planes. As the first plane passes through $OX$ then $n_1$ will be parallel to the $yz-$plane, and similarly $n_2$ will be parallel to the $xz-$plane. Thus: $$ n_1=(0,a,b);\quad n_2=(c,0,d); $$ where: $$ a^2+...


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The tangent to the parabola at the point $(t^2,2t)$ is $x-t^2=t(y-2t)$, because the derivative of $f(y)=y^2/4$ computed at $2t$ is $f'(2t)=t$. Thus $$ ty-x-t^2=0 $$ Such a line must be tangent to the circle, so $$ \frac{|t\cdot0-0-t^2|}{\sqrt{1+t^2}}=\frac{1}{\sqrt{2}} $$ that becomes $$ 2t^4=1+t^2 $$ giving $t^2=1$. Thus the common tangents are $$ y-x-1=0 $$...


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I do not know if it is really helpful, but the center $P$ of the circle has (horrible) coordinates $\dfrac{1}{2(a^2b^2+b^2c^2+c^2a^2)}\begin{pmatrix} a^3(b^2+c^2) \\ b^3(c^2+a^2) \\ c^3(a^2+b^2) \end{pmatrix}$. Indeed, in the plane, we want to find the circle passing through the vertices of the triangle $ABC$. The center $P$ of this circle is the ...


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We can imagine that all the conic sections can be divided into:- (1) Pair of straight lines; (2) Ellipses; (3) Hyperbolas and (4) Parabolas. However, the set of Parabolas can further be subdivided into another subset whose axes of symmetries are only vertical. Quadratic function belongs to this category. Added:-


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For $U(x,y) = Ax ^ 2 + Bxy + Cy ^ 2 + Dx + Ey + F = 0$ the center is located at $$ \left. \begin{align} \frac{\partial U}{\partial x} & = 0 \\\frac{\partial U}{\partial y} & = 0 \end{align} \right\} \begin{aligned} x_c & = \frac{B E - 2 C D}{4 A C - B^2} \\ y_c &= \frac{B D -2 A E}{4 A C - B^2} \end{aligned} $$ This transforms the equation ...


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8,8 and 2,2 are the endpoints of the major axis. The center is at (5,5) the length major axis is 6 sqrt 2 The angle of rotation is 45 degrees. Using the substitution u = x+y, v = x-y Our points in u,v space are (16,0),(11,3),(9,3),(9,-3),(4,0) $\frac{(u-5)^2}{6^2} + \frac{v^2}{b^2} = 1\\ \frac {1}{36} + \frac{v^2}{b^2} = 1\\ b = \frac{18}{\sqrt{35}}$ ...


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Let $P = (h,k)$ be a point on the circle $x^2 + y^2 = \dfrac 12$. $\left( \text{Then}\; h^2 + k^2 = \dfrac 12 \right)$. The origin, $O = (0,0)$ is the center of the circle. So the equation of the line $\overleftrightarrow{OP}$ is $kx - hy = 0$ A line tangent to the line $kx - hy = 0$ must have an equation of the form $hx + ky = C$ for some number $C$. Since ...


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$$\begin{align} \text{Parabola}:\qquad\qquad \qquad \qquad\quad y^2&=4x\\ \frac {dy}{dx}&=\frac 2y=\frac 1a&&\text{at }P(a,2a)\\ \text{Tangent at }P:\qquad\qquad\qquad y-2a&=\frac 1a(x-a)\\ x-ay+a(2a-1)&=0 \end{align}$$ For this line to also be a tangent to the circle $x^2+y^2=\frac12$ (centre $(0,0)$, radius $\frac 1{\sqrt 2}$), the ...


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It is useful to make a sketch to see what is going on. Added: Nice picture, you can see that there are two common tangent lines, that are symmetrical about the $x$-axis. Let $(a,b)$ be the point of tangency to the circle. Then the tangent line has equation $ax+by=1/2$. To find the point(s) of intersection of this tangent line with the parabola, we solve $y^...


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I may be misunderstanding something, but it seems fairly obvious to me that the best circle is of diameter 3 cm. Here's my argument: Start with a circle with diameter 0, and slowly increase it to a diameter of 1 cm. During this time, the circle only intersects with the inner ellipse. Thus, the "weighted area" (with the pluses and minuses as stated) goes ...


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Let us rotate the second figure so that $AC$ is horizontal. Now $EF$ is a tangent line to the lower half of the ellipse. At this point the equation of the ellipse is: $$\left(\dfrac{x}{30}\right)^2 + \left(\dfrac{y}{1.5}\right)^2 = 1 $$ And the function for the lower half of it is: $$y = -1.5\sqrt{1 - \left(\dfrac{x}{30}\right)^2} $$ Recall that the ...


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Consider the two points at which the tangent planes touch the paraboloid: $\left(x_1, y_1,\frac{x_1^2+y_1^2}{2}\right)$ and $\left(x_2, y_2,\frac{x_2^2+y_2^2}{2}\right)$ The tangent planes to these points are: $z-\frac{x_1^2+y_1^2}{2}=x_1(x-x_1)+y_1(y-y_1)$ and $z-\frac{x_2^2+y_2^2}{2}=x_2(x-x_2)+y_2(y-y_2)$. (This is from the equation $z-z_0=f_x(x_0,...


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Let the tangent planes touch the surface at $(x', y',z')$ and $(x'',y'',z'')$, $$\left \{ \begin{array}{rcl} x'x+y'y &=& z+z' \\ x''x+y''y &=& z+z'' \end{array} \right.$$ The line of intersection $(\xi, \eta, \zeta)$ is $$(\xi, \eta, \zeta)= \left( \frac{\begin{vmatrix} \zeta+z' & y' \\ \zeta+z'' & y'' \end{vmatrix}} ...


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Let $H$ be the maximum height achieved. Imagine that you are viewing the tennis game upside down. Let the origin $O (0,0)$ be the highest (lowest when upside down) point of the ball. The trajectory of the ball is then a parabola opening upwards with equation $y=ax^2$ and passes through $(7,H-3)$ and $(11,H)$, i.e. $$\begin{align}\begin{cases}H-3&=a(...


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Draw a circle radius $r$ centre the origin. Its equation is $x^2+y^2=r^2\ (1)$. If it intersects the hyperbola at $(x,y)$ then we have $ar^2+2bxy=c$, so $x^2y^2=\left(\frac{c-ar^2}{2b}\right)^2\ (2)$. Since $x^2,y^2$ are positive numbers we can apply the arithmetic/geometric mean result to (1) and (2) to get $\left(\frac{r^2}{2}\right)^2\ge\left(\frac{c-ar^2}...


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First of all, $x=0$ is a common tangent. Let $(p,q)$ be the tanget point on the circle, and let $(s,t)$ be the tangent point on the parabola. Here, $p,q,s,t\not=0$. The equation of the tangent line at $(p,q)$ is given by $$(p-3)(x-3)+qy=9\implies y=\frac{3-p}{q}x+\frac{3p}{q}$$ Also, the equation of the tangent line at $(s,t)$ is given by $$y-t=\frac{2}{t}(...


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hint : take tangent equation of parabola and find it's perpendicular distance from point $(3,0)$ and equate it to $3$ . https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line


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Consider the equation $x_0^2 - x_1^2 = rx_2^2$ (visualize this as a cone in any way you choose) and then let $r\to 0$ to see how the cone degenerates into two intersecting planes.


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"Subtends right angle"? Do you mean a line segment AB such that A and B lie on the parabola and angle AOB is a right angle? IF the A and B are $(x, 2\sqrt{x})$ and $(x,-2\sqrt{x})$ Then the length of the line segment is, of course, just the difference in y values, $4\sqrt{x}$. That would be the case when the two sides of the angle make 45 degree angles ...


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The general equation of a parabola is $$ f(x) = a x^2 + bx + c $$ it has three parameters. We know $f(11) = 0$, so we have $$ f(11) = 0 $$ we also know $$ 3 = f(11-4) = f(7) $$ further we lobbed from ground level, such that we are 22m away, so this means $$ 0 = f(-11) = a 11^2 - 11 b + c $$ This gives the linear system in the unknowns $a,b,c$: $$ \left[ \...


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The question can be solved mentally with a bit of creativity. Setting $x=0, y = h$ for the vertex, the ball travels $7$ ft horizontally to drop to a height of $3$ ft, and a further $4$ ft horizontally to drop to the ground. Since the horizontal velocity of a projectile (w/o resistance) is constant, the ratios of the distances exactly depict the ratio of ...


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First in order to have an equation you must have a coordinate system! I suspect that you are taking the position from which the ball is lobbed to be then "0" point, you want the parabola to pass through (0, 0) and (22, 0). You want the y value, when x= 22- 4= 18, to be at least 3. Writing the parabola as $y= ax^2+ bx+ c$ we must have (a) $0= a(0^2)+ b(0)+ ...


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As far as your diagram looks, you might try some ombination of parabolas to fit your function.. Your lower part (both sides) look like more or less from a same parabola, andthe upper part from a different one. So you might make a function like $f(x)$=Parabola A when y coordinates below a certain point and Parabola B otherwise. Now what these two (or ...


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Some representations of the ellipsoid: 1. $$ (u - u_c)^\top A (u - u_c) = 1 $$ where $u = (x,y,z)^\top$ and $u_c = (x_c, y_c, z_c)^\top$ and $A$ is a definite matrix with eigen values $r_a^{-2}$, $r_b^{-2}$ and $r_c^{-2}$ 2. $$ u = u_c + u_x x + u_y y + u_z z $$ where $A = (u_x, u_y, u_z)$ is a regular $3\times 3$ matrix. Where one can choose $u_x$, $u_y$...


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Okay then. One way to achieve what you want is via the following: First use a combination of $x$ and $y$ rotations to align the $x$ axis with the minor axis of the ellipse. Then rotate about the $x$-axis by angle $\phi$. (Remark: the following paragraph is new) To find out the angle $\phi$, we start by thinking about what happens if we wrote a flat ...


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Assume a point $(u,v)$ on the parabola, because the derivative of function $y=x^2+1$ is $y' = 2x$. Therefore the tangent line passing through $(u,v)$ has equation: $y - v = 2u(x-u)$...(1) (i.e. statement in the text: ($xx_1=2a(y+y_1)$) ) In the same way, the norm line passing through $(u,v)$ has equation: $ y -v = - 1/(2u) (x-u)$...(2) Setting $y = 0$ in ...


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let L and H be the length and breadth of the required rectangle respectively $$\frac{(L/2)^2}{a^2}+\frac{(H/2)^2}{b^2}=1$$ $$\frac{(L)^2}{4a^2}+\frac{(H)^2}{4b^2}=1$$ $$H=\frac{b}{a}\sqrt{4a^2-L^2}$$ Area=L*H $$A= L*\frac{b}{a}\sqrt{4a^2-L^2}$$ $$\frac{dA}{dL} =\frac{b}{a}\sqrt{4a^2-L^2}-\frac{L^2b}{a\sqrt{4a^2-L^2}}=0$$ $$\frac{b*(4a^2-2L^2)}{a\sqrt{...


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Apply the linear transformation $(x, y) \mapsto (ax, by)$. The transforms your ellipse to the unit circle. It also transforms parallel lines to parallel lines, and midpoints to midpoints. Thus if you can prove the statement for the unit circle, you're done. But for the unit circle, the chords orthogonal to the vector $(p, q)$ have midpoints on the line $-qx +...


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$$\Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + a^2(c^2-b^2) = 0 $$ By the midpoint of a parabola, the $x$-value of the midpoint of the intersection of the line and the ellipse will be at $$\frac{-2mca^2}{2(b^2+a^2m^2)} = \frac{-mca^2}{b^2+a^2m^2}$$ Note that the $y$ value of the midpoint is linearly dependent on the $x$ value of the midpoint $(y=mx+c)$. ...



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