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10

This picture might be a good supplement to Neal's explanation, which is spot-on. If you imagine that your flashlight is pointing straight down from the tip of the cone, then the wall is any one of the colored areas, depending on the angle you hold it at.


6

Start with the flashlight parallel to the wall. If you incline it slightly away from the wall, the (conical) beam will still hit the wall and you'll get a hyperbola. It is only after the angle of the flashlight to the wall is greater than the angle of the light cone that the hyperbola vanishes at infinity. (In practice, of course, it is rather before then ...


6

We can write the quadratic part as follows: $$13x^2 - 20xy + 52y^2 = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}13 & -10 \\ -10 & 52\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$$ By choosing a change of coordinates that diagonalizes this matrix, we can transform any equation like this one into an ellipse with axes parallel to the ...


4

I offer the above picture as a proof without words to show that, even if the afore-mentioned ellipse were uniquely determined in terms of size and dimensions $($and I'm not saying that it is$)$, it cannot be uniquely determined in terms of position. Ellipse #$2$ is the reflection of Ellipse #$1$ with regard to $AB$, whereas Ellipse #$3$ and Ellipse #$4$ ...


4

First attempt: Going from a sketch in GeoGebra the equation is roughly $$ \left(\frac{x - 513 }{513}\right)^2 + \left(\frac{y- 100}{877.5}\right)^2 = 1 $$ Algebraic model: Assuming the minor axis is parallel to the $x$-axis, the equation of the ellipse is $$ \left(\frac{x - b}{b}\right)^2 + \left(\frac{y - 100}{a}\right)^2 = 1 \quad (1) $$ with unknowns ...


4

As @AchilleHui mentions, the midpoints of two parallel chords lead to the point of tangency ($T$) with a third parallel line. Note that the line of midpoints is parallel to the axis of the parabola. By the reflection property of conics, the line of midpoints and the line $\overleftrightarrow{TF}$ make congruent angles with the tangent line. So, two sets ...


4

Take two parallel lines $l_1,l_2$ cutting the ellipse in $A_1,B_1,A_2,B_2$. Let $C_i$ be the midpoint of $A_i B_i$: the center of the ellipse lies on the $C_1 C_2$ line. Once the center $O$ of the ellipse is found, take a circle with center $O$ cutting the ellipse in the vertices of a rectangle: the symmetry axis of the rectangle are the axis of the ellipse ...


4

Let three tangents to a parabola form a triangle. Then Lambert's theorem states that the focus of the parabola lies on the circumcircle of the triangle. So draw three tangents. Find the triangle that they form and construct the circumcircle. Start again with two of the original tangents and a new tangent. Find the triangle they form and construct the ...


3

A homogeneous equation, ie, an equation with all terms of same degree, will always represent a set of straight lines passing through the origin. With this fact in mind, consider the following example: I am given a curve of the form $ax^2+by^2+cxy+dy+ex+f=0$ and the line $px+qy=r$ which intersects it at points $A$ and $B$. I need to find the joint ...


3

First of all, the curve is clearly a non-degenerate conic or else we won't be finding its area. In the equation of a conic if the coefficient of $xy$ is $2h$, of $x^2$ is $p$ and of $y^2 $ is $q$ then the rule is that if $h^2<pq$, then it is an ellipse, which it is. The centre of a conic represented by the equation $ax^2+by^2+2hxy+2gx+2fy+c=0$ is given ...


3

This is a complete proof, and it might get a bit repetitive because I'm using the Sine Law again and again. I'm trying to find something more elegant, but this is the only thing thing I could come up with right now.( I've used a different diagram because my internet wasn't working while I was working this out. Just skip over to the ending note if you just ...


2

If you look closely at W|A's graph you will see that it is not actually centered at the origin, just close to it. If it were centered at the origin it would be symmetric with respect to the origin, and replacing $x$ with $-x$ and $y$ with $-y$ would give the same equation, but it does not. I can see two ways to solve this problem, non-ad-hoc. First, do the ...


2

I am interpreting the question as asking for a minimum number of points that determine a unique ellipse passing through them. Four points is not enough. The two ellipses $$ x^2+2y^2=3\qquad\text{and}\qquad 2x^2+y^2=3 $$ both pass through the points $(x,y)=(\pm1,\pm1)$ (all four sign combinations). Five points $P_j=(x_j,y_j), j=1,2,3,4,5,$ (in general) does ...


2

$\rm PF.PG$ = Power of P wrt the circle with CG as diameter, the equation of whose is: $$x(x-(1/a)(a^2+b^2)\sec\theta)+y^2=0$$ So, $${\rm PF.PG}=|a\sec\theta(a\sec\theta-(1/a)(a^2+b^2)\sec\theta)+b^2\tan^2\theta|=b^2$$ Similiarly: $\rm PF.PG'$ = Power of P wrt the circle with CG' as diameter, the equation of whose is: $$x^2+y(y-(1/b)(a^2+b^2)\tan\theta)=0$$ ...


2

Your assumptions are half correct. Here's why: The apex can never be lower than the higher point of the starting and endpoints, The best you can get is, the highest point of the two (start or end) itself will become the apex. You can always define a vertical parabola $y=ax^2+bx+c$ passing through $3$ given points. Just substitute the points' coordinates in ...


2

Unlike the one I gave in this, this is a "pure geometric solution". The figure is pretty much self-explanatory but still I will spell it out: $P$ and $Q$ are points on the ellipse. $PR$ and $QR$ are the tangents at these points. $PC$ and $QC$ are the normals at these points. Line $AB$ is a directrix. $F_1$ is a focus. $CD$ is drawn parallel to the major ...


1

Referring to G-man's sketch, $ \epsilon = eccentricity, RQB = \phi , F_1QR =\psi, then \frac {\cos\psi }{\cos\phi} = \epsilon $ The proof is very near from here.


1

This exercise can be understood as an application of a general result about perimeter bisectors of triangles. Proposition. Given $\triangle ABC$ with incircle $\bigcirc I$ meeting the edges at $D$, $E$, $F$ as shown. If $F^\prime$ is the point opposite $F$ in $\bigcirc I$, and if $F^{\prime\prime}$ is the point where $\overleftrightarrow{CF^\prime}$ ...


1

By Newton's second law $F = ma = m(0,-g)$, which es equivalent to the set of equations: \begin{align} \ddot x(t) &= 0\\ \ddot y(t) &= -g. \end{align} After integrating these, we get that: \begin{align} x(t) &= x_0+v_{x}t\\ y(t) &= y_0+v_yt-\frac{gt^2}{2}, \end{align} where $(x_0,y_0)$ is the initial position of the object and $(v_x,v_y)$ its ...


1

Hints and References : Look up the detailed guide here. The parabola is shifted and it is rotated by 180 degrees . ( The -ve sign of p indicates that it has been flipped) This might also help.


1

Yes, this is a parabola. You can factor out like this: $$(y-3)^2 = -12(x+3)$$ This means, the vertex is located at $(-3,3)$. The parabola is opening towards the left. In this case, $-4p = -12$, $p=3$. This means the distance from focus to vertex is 3. If you draw the parabola, the focus should be located at $(-6,3)$. The directrix should be a vertical ...


1

The wall is rectangular, but one is interested only in a semicircular part of it. The force on a horizontal slice of $dA = 2x\times \,dy$ according to Pascal's law is: $$ dF = p \,dA = \rho g h 2 x \,dy $$ All points on a circle with radius $5$ and center at the origin fulfill $x^2+y^2=25$. It is used here to express $x$ in terms of $y$ and leads to $$ dF ...


1

This isn't a "pure geometric solution" since I've used some results from analytic geometry, but it does save you all the hassle of finding co-ordinates and calculating slopes and distances. All the $\color{green}{green}$ angles in the figure are equal to $\phi$ , where $\tan\phi={b\over a}\csc\theta$. If you use a trigonometric identity you'll get ...


1

Your ellipse is centered at the origin. Therefore let's look at the function $$R(t):=y^2(t)+z^2(t)=(a^2+c^2)\cos^2 t+(b^2+d^2)\sin^2 t +2(ab+cd)\cos t\sin t$$ that represents the squared distance of the moving point from the origin. It can be rewritten as $$R(t)={1\over2}\biggl(a^2+c^2+b^2+d^2 +\bigl((a^2+c^2)-(b^2+d^2)\bigr)\cos(2t)+2(ab+cd)\sin(2t)\biggr)\ ...


1

This is the Desmos graph of the equation from the math question.


1

No, the ellipse as described is not uniquely specified. For your example, any ellipse in the family $$ \frac{(x-y)^2}{a^2} - \frac{(x+y)^2}{b^2} =1 $$ with $$\frac{1}{a^2} +\frac{1}{b^2} = \frac{1}{5^2} $$ will pass thru $(5,0)$ and $(0,5)$ One of these is a nice circle of radius $5$, another (with $a = \frac{25}{3}, b=\frac{25}{4}$) is an ellipse with ...


1

Suppose the original conic section is $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ The tangent line at $(x_0,y_0)$ is $$y-y_0=m(x-x_0)$$ To find $m$, take derivative on both sides of the conic section: $$2Ax+By+Bx\frac{dy}{dx}++2Cy\frac{dy}{dx}+D+E\frac{dy}{dx}=0$$ Solve for $\frac{dy}{dx}$ and plug into $m$: $$y-y_0=\frac{-2Ax_0-By_0-D}{Bx_0+2Cy_0+E}(x-x_0)$$ ...


1

Here's a way to think of it. If you shift the point A (and the rest of the curve, with it's tangent line) to the origin you can look at the linear part at the origin (tangent line), then shift that line back to A. First shift A to (0,0): $(x+x_A)^2+(y+y_A)^2+6(x+x_A)-8=x^2+2x\cdot x_A+x_A^2+y^2+2y\cdot y_A+y_A^2+6x+6x_A-8$ Now look at the linear part, and ...


1

Note that you can rewrite this, with a little work, as $$4(x+2y+1)^2+9(-x+2y+1)^2=576=24^2$$ If we put $X=x+2y$ and $Y=-x+2y$ the determinant of the transformation is $4$ and we get $$\frac {(X+1)^2}{12^2}+\frac {(Y+1)^2}{8^2}=1$$ as a transformed ellipse with its axes parallel to the new co-ordinate axes. The new ellipse has major/minor axes of lengths ...


1

Plot any point $(x_1,y_1)$ in the plane. Then reflect that point across the line $y=-x$ and see where the reflection of the point is. The reflection will be at $(x_1',y_1') = (-y_1,-x_1)$. Note that not the signs of both coordinates are changed, not just the $x$ coordinate. The result (with the correct signs of both coordinates) is that the reflected ...



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