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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 +... 4 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, $$\alpha+\beta=\frac{2ab^2}{a^2+b^2},\quad \alpha\beta=\frac{a^2(... 3 It is useful to make a sketch to see what is going on. Added: Nice picture, you can see that there are two common tangent lines, that are symmetrical about the x-axis. Let (a,b) be the point of tangency to the circle. Then the tangent line has equation ax+by=1/2. To find the point(s) of intersection of this tangent line with the parabola, we solve y^... 3 Consider the equation x_0^2 - x_1^2 = rx_2^2 (visualize this as a cone in any way you choose) and then let r\to 0 to see how the cone degenerates into two intersecting planes. 2 The question can be solved mentally with a bit of creativity. Setting x=0, y = h for the vertex, the ball travels 7 ft horizontally to drop to a height of 3 ft, and a further 4 ft horizontally to drop to the ground. Since the horizontal velocity of a projectile (w/o resistance) is constant, the ratios of the distances exactly depict the ratio of ... 2 The generator line of the first cone is \frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}. Setting y=0 gives$$\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}... 2 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}...

2

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 ...

2

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 ...

2

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 ...

1

[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 ...

1

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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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 $t = 0$ ...

1

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: $$\sqrt{\left(\frac{2a\sqrt{r^2(a^2+b^2)-b^2}}{a^2+b^2}\right)^2+\left(\frac{2b\sqrt{r^2(a^2+b^2)-a^2}}{a^2+b^2}\... 1 The line of intersection of the planes through OX and OY does not lie on the xy-plane. Let n_1 and n_2 be the unit vectors perpendicular to the planes. As the first plane passes through OX then n_1 will be parallel to the yz-plane, and similarly n_2 will be parallel to the xz-plane. Thus:$$ n_1=(0,a,b);\quad n_2=(c,0,d); $$where:$$ a^2+...

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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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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 $$(X-a)^2+Y^2+\left(C\left(1-\frac Xa-\frac Yb\right)\right)^2=X^2+(Y-b)^2+\left(C\left(1-\frac Xa-\frac Yb\right)\right)^2=X^2+Y^2+\left(C\left(1-\frac Xa-\frac Yb\right)-C\right)^2$$ ...

1

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 ...

1

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 ...

1

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) $\frac{(u-5)^2}{6^2} + \frac{v^2}{b^2} = 1\\ \frac {1}{36} + \frac{v^2}{b^2} = 1\\ b = \frac{18}{\sqrt{35}}$ ...

1

\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 ...

1

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 ...

1

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 $z-z_0=f_x(x_0,... 1 Let the tangent planes touch the surface at$(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 $$(\xi, \eta, \zeta)= \left( \frac{\begin{vmatrix} \zeta+z' & y' \\ \zeta+z'' & y'' \end{vmatrix}} ... 1 First in order to have an equation you must have a coordinate system! I suspect that you are taking the position from which the ball is lobbed to be then "0" point, you want the parabola to pass through (0, 0) and (22, 0). You want the y value, when x= 22- 4= 18, to be at least 3. Writing the parabola as y= ax^2+ bx+ c we must have (a) 0= a(0^2)+ b(0)+ ... 1 As far as your diagram looks, you might try some ombination of parabolas to fit your function.. Your lower part (both sides) look like more or less from a same parabola, andthe upper part from a different one. So you might make a function like f(x)=Parabola A when y coordinates below a certain point and Parabola B otherwise. Now what these two (or ... 1 Some representations of the ellipsoid: 1.$$ (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... 1 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 ... 1 Continuing the computation: the x-value of the midpoint isx_{mid} = \frac{x_1+x_2}{2} = -\frac{mca^2}{b^2+a^2m^2}$$and the corresponding y-value is$$y_{mid} = mx_{mid}+c = \frac{b^2c}{b^2+a^2m^2}. $$Therefore,$$ma^2 y_{mid} + b^2 x_{mid} = 0$and that is an equation of a straight line through the origin. Remark: that line is the polar ... 1 A generic line intersects a conic in$2$points. If the conic and the line are rational (given by rational coefficients), then the set consisting of those two points is invariant by$\Bbb Q$-automorphisms, so either each point is rational, either they are defined over a quadratic extension of$\Bbb Q\$ and the automorphism of that extension swaps the two ...

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