I am trying to determine the equation of directrix and focus of the parabola which has tangents $y=x$ at $(3,3)$ and $y=-x$ at $(1,-1)$. Drawing a rough picture suggest that the parabola is oblique. Also since the foot of perpendicular from focus to any tangent lies on the tangent at vertex, this implies that the line joining the focus and the foot of perpendicular to tangent $y=x$ is of the form $x+y=k_1, k_1\in\mathbb{R}$ and also line joining focus and foot of perpendicular to tangent $y=-x$ is of the form $y=x+k_2,k_2\in\mathbb{R}$.  One standard method is to consider $ax^2+by^2+2gx+2fy+2hxy+c=0$ and write the equation of tangents with point of contact $(x_1,y_1)$ as $axx_1+byy_1+g(x+x_1)+f(y+y_1)+h(x_1y+xy_1)+c=0$ for both the tangents and compare with the equation of tangents but that is rather tedious.   It is unclear on how to proceed from here. Any hints involving only the geometry of the standard parabola without involvement of linear algebra techniques are appreciated. Thanks.
121464
If a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$, then find the focus, axis of the parabola and its directrix.   What I thought: Since the 2 tangents are perpendicular,the origin must lie on the directrix and the line joining A and B is a focal chord. Don't know how to proceed from here..
88670
If   two points $P$ and $Q$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ whose center is $C$ be such that $CP$ is perpendicular to $CQ$ and $a&lt;b$,then prove that $\frac{1}{CP^2}+\frac{1}{CQ^2}=\frac{1}{a^2}-\frac{1}{b^2}$.    I let $P(a\sec\alpha,b\tan\alpha)$ and $Q(a\sec\beta,b\tan\beta)$,Center $C$ is $(0,0)$. $CP^2=a^2\sec^2\alpha+b^2\tan^2\alpha$ and $CQ^2=a^2\sec^2\beta+b^2\tan^2\beta$ We need to prove that $\frac{1}{a^2\sec^2\alpha+b^2\tan^2\alpha}+\frac{1}{a^2\sec^2\beta+b^2\tan^2\beta}=\frac{1}{a^2}-\frac{1}{b^2}..........(1)$ We are given that slope of $CQ\times$ slope of $CP=-1$ $\frac{b\tan\beta}{a\sec\beta}\times\frac{b\tan\alpha}{a\sec\alpha}=-1..........(2)$ I am not able to prove equation $(1)$ from the given equation $(2.)$
16760
Let's take a point $E = (x_A, y_A)$ and line $p: Ax+By+C=0$ Let's find the reflection point of point $E$ over the line $p$    I find the equation of a line perpendicular to $p$ and passing through point E I find the distance between $E$ and $p$: $d(E,p)$ I solve the equation $$d(E,p)=\frac{|Ax_A +B(x_A+m)+C|}{\sqrt{A^2+B^2}}$$   Is there a more effective way to reflect a point around a line?
61841
I need a synthetic proof on this problem without the use of trigonometry.     Question: Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AF\perp BE$.    The proof is very easy using coordinate geometry.    I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.
30823
Find the equations of the tangent line(s) to $\;f(x) = x^3 -4x^2 + 2\;$ that are parallel to the line $y= -4x +3$.  I know the slope is -4 and $f'(x)= 3x^2-8x.$
66865
Given four points $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$,$(x_4,y_4)$. How can we efficiently test them to make sure whether they are vertices of a parallelogram?  I think evaluating and comparing the distance between two points seems to be inefficient. What do you think? Do you have any better method?
41909
Find the condition on $a$ and $b$ so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by.$  The required condition is $a^2&gt;8b^2$.I dont know how to prove it.I tried.  Let $(h,k)$ be the point from where tangents are drawn to the parabola $y^2=4ax$.Let $m_1,m_2$ be the slopes of the tangents.Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$ and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.  As $y=m_1x+\frac{a}{m_1}$ and $y=m_1x-2bm_1-bm^3_1$  are the same lines.Therefore,$\frac{a}{m_1}=-2bm_1-bm^3_1$  $bm^4+2bm^2+a=0$  And I am stuck.I cannot get the desired condition.Please help me.
88664
Let $L$ be the line given by the equation $4x − 3y = 0$. Let $S : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be reflection through that line, and let $P : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be projection onto that line.  Determine, geometrically without doing any computations, whether there exist non-zero vectors $x$ such that: $$ a) \qquad S(x) = 1 · x $$ $$ b)\qquad P(x) = 1 · x $$ Can someone tell me how to solve this? Does anyone know what kind of arithmatic this is? I wanna look it up on google but I don't know what it's called. Looks a bit like eigenvectors to me but I'm not entirely sure...  All help is appreciated.
57947
I have two quadratic bezier curves and I want to find whether they have a common tangent, and if so where it is.  If the first bezier is defined by points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, then the points on the curve have coordinates:  $x = (x_1 - 2x_2 + x_3)t^2 + (2x_2 - 2x_1)t + x_1 \\ y = (y_1 - 2y_2 + y_3)t^2 + (2y_2 - 2y_1)t + y_1$  Or to simplify things:  $x = A_xt^2 + B_xt + C_x \\ y = A_yt^2 + B_yt + C_y$  Likewise, I can get an equation for the second bezier:  $x = D_xs^2 + E_xs + F_x \\ y = D_ys^2 + E_ys + F_y$  The gradient of the tangent, $m$, to the first curve at $t$ is $\dfrac{2A_yt + B_y}{2A_xt + B_x}$,  which must be equal to the tangent to the second curve at $s$, $\dfrac{2D_ys + E_y}{2D_xs + E_x}$.  So $m = \dfrac{2A_yt + B_y}{2A_xt + B_x} = \dfrac{2D_ys + E_y}{2D_xs + E_x}$  Which I rearranged to get:  $t = \dfrac{2(B_yD_x - D_yB_x)s + B_yE_x - B_xE_y}{4(A_xD_y - A_yD_x)s + 2(A_xE_y - A_yE_x)}$  The gradient of the tangent is also equal to the difference in the y-coordinates over the difference in x-coordinates:  $m = \dfrac{D_ys^2 + E_ys + F_y - (A_yt^2 + B_yt + C_y)}{D_xs^2 + E_xs + F_x - (A_xt^2 + B_xt + C_x)}$  But that's as far as I got. The algebra looks too messy from here. Can anyone solve this, or is there a better approach?
35316
I have the following problem that I don't know how to solve:  Let $\textbf{r}(t)$ be a curve and let $r(t):= |\textbf{r}(t)|$. Let the curve $\textbf{r}(t)$ be defined by: $$\textbf{a}(t)=\frac{GM}{(r(t))^3}\textbf{r}(t)$$ Where $\textbf{a}(t) = \textbf{r}''(t)$ is the acceleration of the particle and $G,M \in \mathbb{R}$. Suppose that $\textbf{r}(t)$ and $\textbf{r}'(t)$ are not parallel. The objective is to determine the trajectory $\textbf{r}(t)$ in polar coordinates.  So basically we have the following equation: $$\textbf{r}''(t)=\frac{GM}{(r(t))^3}\textbf{r}(t)$$ The thing is that I don't know how I should proceed from now on. If we'd have just normal scalar function this would be a differential equation and I think I would be able to solve it, but like this I have no idea how to approach the problem. How can I solve this?  Edit: If $$\textbf{r}(t):=(x(t),y(t))$$ then the equation turns into the following system of equations: $$x''=\frac{GM}{(\sqrt{x^2 + y^2})^3}x$$ $$y''=\frac{GM}{(\sqrt{x^2 + y^2})^3}y$$ I don't know if this helps but I think that this turns thing easier perhaps.
68925
I'm having trouble understanding the solution presented below (It's from a textbook). I tried to get something similar, but to no avail. Help me find the way he derived those a,b,x2 and y2 expressions.   Point p is a point which we want to reflect (It's a simple structure with two integers x and y representing coordinates on a 2d plane); (x0,y0) and (x1,y1) are coordinates of endpoints of a line.  Expected result is a new point with reflected coordinates.   Point mirror(Point p, int x0, int y0, int x1, int y1) {     double dx,dy,a,b;    long x2,y2;    Point p1; //reflected point to be returned      dx  = (double) (x1 - x0);    dy  = (double (y1 - y0);     a   = (dx * dx - dy * dy) / (dx * dx + dy*dy);    b   = 2 * dx * dy / (dx*dx + dy*dy);     x2  = Math.round(a * (p.x - x0) + b*(p.y - y0) + x0);     y2  = Math.round(b * (p.x - x0) - a*(p.y - y0) + y0);     p1 = Point((int)x2,(int)y2);      return p1;  }
102651
There are 2 questions(part of same question but I divided it into two):    Q1. Prove that the length of the focal chord of the ellipse $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ which is inclined to the major axis at an angle $\theta$ is $$\frac {2ab^2}{a^2\sin^2\theta+b^2\cos^2\theta}$$ I tried to solve this using the parametric form of a line, i.e., $(x,y)=(ae+r\cos\theta,r\sin\theta)$, plugging this into the given equation to find $r_1-r_2$ which is giving a different solution.    Q2. If $PSQ$ and $PS'R$ are two chords of an ellipse through its two focii S and S', then prove that $$\frac {PS}{SQ}+\frac {PS'}{S'R}=2\left(\frac {1+e^2}{1-e^2}\right)$$ Here I fixed $P \equiv (a\cos\theta,b\sin\theta)$ and found out the parametric angles of $Q$ and $R$ using the result $$\tan\theta_1/2*\tan\theta_2/2=\left(\frac {e-1}{e+1}\right)^{\pm1}$$, $+$ when chord passes through $S$ and $-$ when chord passes through $S'$. But it gives a very ugly expression. So how do I do it  in an easy way? Hints are welcome.
46437
I have come across this question that I need a tip for.     Find the equation (general form) of the plane passing through the point $P(3,1,6)$ that is orthogonal to the vector $v=(1,7,-2)$.   I would be able to do this if it said "parallel to the vector"  I would set the equation up as $(x,y,x) = (3,1,6) + t(1,7,-2)$ and go from there.  I don't get where I can get an orthogonal vector. Normally when I am finding an orthogonal vector I have two other vectors and do the cross product to find it.  I am thinking somehow I have to get three points on the plane, but I'm not sure how to go about doing that.  Any pointers?  thanks in advance.
65812
I'm trying to compute the shadow of 3D objects on the ground. To do this I'm assuming parallel rays from Sun origin.   Let's assume the Sun direction is given by azimuth $\varphi$ and elevation $\theta$ angles. So the direction unit vector is retrieved as:  $$ \hat{\mathbf{s}} = \mathbf{R}_y\left(\theta\right)\mathbf{R}_z\left(\varphi\right)\hat{\boldsymbol{\imath}}$$  where $\mathbf{R}_{x/y/z}$ are the rotation matrices along the $x, y, z$ axes given by, for example, Euler-Rodrigues equation, and $\hat{\boldsymbol{\imath}}$ is the unit vector along $x$-axis.  A point on a ray-line has the equation:  \begin{equation} \mathbf{r} = \mathbf{r}_p + \lambda\hat{\mathbf{s}}  \tag{1} \end{equation}  being $\mathbf{r}_p$ a point on that line.  A point on a general plane is given by the equation:  \begin{equation} \left(\mathbf{r} - \mathbf{r_0}\right)\cdot \hat{\mathbf{n}} = 0 \tag{2} \end{equation}  being $\mathbf{r}_0$ a point on the plane and $\hat{\mathbf{n}}$ the unit vector normal to that plane.   Ensuring the validity of both equations (1) and (2) allows to compute the $\lambda$ multiplier and retrieve the position vector of the projected point (writing $\tilde{\mathbf{r}} = \mathbf{r_p} - \mathbf{r_o}$)  \begin{gather} \lambda = - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right) \\ \mathbf{r}_\text{proj} = \mathbf{r}_p - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right)\hat{\mathbf{s}} \tag{3} \end{gather}  Applying eq. (3) to all the vertices of a mesh $\left\{\mathbf{r}_{p_i}\right\}$ .   A problem can be noted: $\hat{\mathbf{s}}\cdot \hat{\mathbf{n}}$ can be zero (e.g. Sun is horizontal, i.e. the ray is ortogonal to the normal of plane).   Are there some conceptual mistakes in my reasoning? Is there a way to rewrite the operation as a linear operator applied to the set of vertices position vectors?   $$ \left[\mathbf{A}\right]\left\{\mathbf{r}_{p_i}\right\} = \mathbf{r}_\text{proj}$$
61033
Consider three skew (non-intersecting) lines $L,M,N$ in $\mathbb{P}^3$. Each line is given by two equations of the form $\alpha_{i,j}^\top z = 0, \, i=1,2,3, j = 1,2$, where $z=(z_0,z_1,z_2,z_3)$ are the homogeneous coordinates of $\mathbb{P}^3$ and $\alpha_{i,j} \in \mathbb{P}^3$.  What i want is to show that up to projective equivalence the equations for the three lines can be written as (this is possible according to the hint of exercise 2.12 in Harris, Algebraic Geometry - a first course)  $L: \, z_0=z_1=0 \\ M: \, z_2=z_3 = 0 \\ N: \, z_0=z_3, \, z_1=z_2.$  Since say $L, M$ are non-intersecting, this implies that $\alpha_{i,j}$ for every $i=1,2, j=1,2$ are in general position. So they are projectively equivalent to $(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)$. This gives us  the description that we want for $L$ and $M$. But how about $N$?  Note here that the most general result that we have is projective equivalence between two sets of 5 points in general position. In the context of the present question, i am concerned about two things:  1) The fact that the lines $L,M,N$ are skew does not imply in general that  every group of 5 points among the $\alpha_{i,j}$ will be in general position.   2) The $\alpha_{i,j}$ are six!  Question: What am i missing here? How can the above two issues be addressed?
54215
Find an equation of the plane containing the point $(1, 1, -1)$ and perpendicular to the line through the points $(2, 0, 1)$ and $(-1, 1, 0)$  This is what I have:  I first find the vector between the points: $\vec{n}^{\ } = (2,0,1) - (-1, 1, 0) = (3, -1, 1)$  Next I find the formula for the plane that is perpendicular to the vector by taking the dot product of the vector and the given point + another point of the plane  $(3, -1, 1) \cdot (1 + x, 1 + y, z -1 )\\ 3 + 3x -1 - y -1 -z\\ 3x - y - z +1 = 0$  I am not at all sure if I followed the correct steps
115277
Question: Tangents drawn from the point $P(1, 8)$ to the circle $x^2 + y^2 -6x -4y -11=0$ touch the circle at the points $A$ and $B$. What is the equation of the circumcircle of the triangle $PAB$?  I tried drawing a diagram, and found the length of the tangent. However, I lost as to where to proceed from there. Please help me!