the only information given is ellipse equation x^2/17+y^2/8=1, first I tried to substitute y in ellipse equation with m(x-a)+b, then I tried to make determinant zero, getting 2nd order equation of am^2+bm+c=0, then I tried to get the determinant of this equation higher than zero, noting c/a= m1.m2 and m1.m2=-1, but I don't know the rest.
80457
For a given ellipse, find the locus of all points P for which the two tangents are perpendicular.   I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof.
34167
In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?
104046
I am very close to the solution I think. Since the circles cross the real axis, I want to find mappings from $z \to w$ such that $1/2 \to m$ and $0 \to -m$ and $-1 \to n$ and $1 \to -n$. Using $$ w = \frac{az+b}{cz+d}$$ we get $$ \frac{a+2b}{c+2d} = -\frac{b}{d}$$ and $$ \frac{b-a}{d-c} = - \frac{a+b}{c+d}$$ but where do we go from here? I am a little lost in the simplifying algebra here to solve for the $a,b,c,d$ in the transformation.
82829
I know that for any nonzero $x,y\in\mathbf{R}$,  $$ax^2+by^2+cxy &gt; 0,$$  where $a,b,c\in\mathbf{R}$. What can I deduce about $a$, $b$, and $c$?    For example, letting $x=1$ and $y=0$, I know that $\boxed{a&gt;0}$. Letting $x=0$ and $y=1$, I know that $\boxed{b&gt;0}$. Letting $x=y=1$, I know that $\boxed{c&gt;-(a+b)}$.  What else can I deduce about $a$, $b$, and $c$? How will I know when it's time to stop looking? (Is that last question even answerable?)
88846
i have to prove:   $$\begin{vmatrix} -2a &amp;a+b  &amp;a+c \\   b+a&amp;-2b  &amp;b+c \\   c+a&amp;c+b  &amp;-2c  \end{vmatrix} = 4(a+b)(b+c)(c+a)$$   I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help?
19918
I'm having troubles solving the following determinant:  $$\left|   \begin{array}{cccc}     a_1 &amp; x &amp; \ldots &amp; x \\    x &amp; a_2 &amp; \ldots &amp; \vdots \\     \vdots &amp; \ldots &amp; \ddots &amp; x \\     x &amp; \ldots &amp; x &amp; a_n \\   \end{array} \right|$$  I have tried coming up with recurrent formula, but there's always one lingering row and column. I also tried finding values of x for which determinant is 0, hoping I could get a nice polynomial solution, with no luck.  By subtracting first row from every other I get somewhat nicer result, but there's still that one row/column which I can't solve. If all other columns are added to the first one and then I subtract the first row, then the column doesn't depend on x anymore, but I haven't found 1..n much more useful.
92383
I find this question very interesting, but I am having trouble figuring out how to approach the problem.   I know that the $\det(A^{-1}) = 1/\det(A)$, but I'm unsure of where to go from here. If someone can send me in the right direction, it would be very much appreciated.   Thanks in advance,  Tommy.
118135
We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:  $$\begin{align}3x+2y&amp;=36 \tag1\\ 5x+4y&amp;=64\tag2\end{align}$$  I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$. I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &amp;= 36 \tag{1a}\\ 5x + 5y - y &amp;= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$. I can even use matrices! $(1)$ and $(2)$ could be written in matrix form: $$\begin{align}\begin{bmatrix} 3 &amp;2 \\ 5 &amp;4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&amp;=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &amp;= {\begin{bmatrix} 3 &amp;2 \\ 5 &amp;4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &amp;= \frac{1}{2}\begin{bmatrix}4 &amp;-2 \\ -5 &amp;3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &amp;=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &amp;= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&amp;=8 \\ \therefore y&amp;= 6\end{align}$$  Question Are there any other methods to solve for both $x$ and $y$?
62448
Good Night ... Here I am editing the question and explaining the purpose of the same ... Mathematics course, and I love this science, I am very curious, and these days, studying a book of calculus found in (book from the library of the University) a sheet full of challenges, to share with a colleague, he gave me this challenge, which he also failed to respond ... I tried, I tried, I used the formulas of quadratic equation, and I've been trying since Friday (here in Brazil is Sunday at 22:33), and could not, and I ask your help, and please believe me, I struggled, tried to make substitutions, tried both, and I'm confused ... I must find the values of $ a, b $ and $ c $?  Consider a parabola whose roots, other are $x'$ and $x''$, and such that $x'=-x''$ where $x'&gt; 0$. Let $$A = (x', 0)$$ $$B = (x'', 0)$$ and $V$ the vertex of the parabola. Suppose that the equation of this parabola is $$y = ax^2 +bx+ c$$. Calculate the value of the discriminant $$\Delta=b^2-4ac$$, so that the triangle is equilateral AVB.
53005
$$\begin{bmatrix} 1235 &amp;2344 &amp;1234 &amp;1990\\ 2124 &amp; 4123&amp; 1990&amp; 3026 \\ 1230 &amp;1234 &amp;9095 &amp;1230\\ 1262 &amp;2312&amp; 2324 &amp;3907  \end{bmatrix}$$  Clearly, its determinant is not zero and, hence, the matrix is invertible.   Is there a more elegant way to do this?  Is there a pattern among these entries?
77302
How to evaluate  $$\det\begin{bmatrix}1+x_1y_1&amp;\cdots &amp; 1+x_1y_n\\ \vdots &amp; \ddots &amp; \vdots \\ 1+x_ny_1 &amp; \cdots &amp; 1+x_ny_n\end{bmatrix}$$?  I tried factorisation but I have no idea. I could not factor anything out... I need a hint to kick off.   Edit: A conjecture due to multilinearity: if we let $v_j=(x_1y_j,\cdots,x_ny_j)^T$, then my guess would be $$\det\begin{bmatrix}1+x_1y_1&amp;\cdots &amp; 1+x_1y_n\\ \vdots &amp; \ddots &amp; \vdots \\ 1+x_ny_1 &amp; \cdots &amp; 1+x_ny_n\end{bmatrix}=\det(1,v_2,\cdots,v_n)+\det(v_1,1,\cdots,v_n)+\cdots+\det(v_1,v_2,\cdots,1)=\color{red}{0}$$ where $1$ denote $(1,\cdots,1)^T\in \Bbb K^n$,$\Bbb K$ the underlying number field.
44146
I am looking for a derivation for the inverse of a 2x2 matrix. I am also wondering why the determinant is involved in the expression. I am familiar with high school maths and linear algebra. If there is an intuitive reason for expression i would also be interested in that.
47280
I am having trouble solving the diophantine equation given in the title. This is how far I came:  We can factor in $\mathbb{Z}[i]$ $y^2+4=x^3\Rightarrow (y+2i)(y-2i)=x^3$. I want to show now that their gcd is a unit so in $\{\pm 1,\pm i\}$. Suppose $d=u+vi$ is the gcd, then $u+vi\text{ | }2y$ and $u+vi\text{ | }4i$. Now taking the norm, we find $u^2+v^2\text{ | }4y^2$ and $u^2+v^2\text{ | }16$. From the second part we find that $d=u+vi\in\{\pm 1,\pm i, \pm 4,\pm 4i,\pm 1\pm i,\pm 2\pm 2i\}$. Obviously $\pm 4$ and $\pm 4i$ cannot divide $y\pm 2i$. I don't know how to show it for the others.   If we assume that $y+2i$ and $y-2i$ are coprime and using the equation $(y+2i)(y-2i)=x^3$ we find that $y-2i=\epsilon \alpha^3$ and $y+2i=\delta \beta^3$ where $\epsilon$ and $\delta$ are units. From here on it is straightforward to find what $\alpha$ and $\beta$ can be. My only problem is showing that $y-2i$ and $y+2i$ are coprime.
83676
I have the following data:-   I have two points ($P_1$, $P_2$) that lie somewhere on the ellipse's circumference. I know the angle ($\alpha$) that the major-axis subtends on x-axis. I have both the radii ($a$ and $b$) of the ellipse.   I now need to find the center of this ellipse. It is known that we can get two possible ellipses using the above data.  I have tried solving this myself but the equation becomes so complex that I always give up.  This is what I have done till now:-  I took the normal ellipse equation $x^2/a^2 + y^2/b^2 = 1$. To compensate for the rotation and translation, I replaced $x$ and $y$ by $x\cos\alpha+y\sin\alpha-h$ and $-x\sin\alpha+y\cos\alpha-k$, respectively. $h$ and $k$ are x and y location of the ellipse's center.  Using these information I ended up with the following eq:- $$a B_1\pm\sqrt{a^2 B_1^2 - C_1(b^2 h^2 - 2 A_1 b^2 h)} = a B_2\pm\sqrt{a^2 B_2^2 - C_2 (b^2 h^2 - 2 A_2 b^2 h)} \quad (1)$$  where $A = x\cos\alpha +y\sin\alpha$, $B = -x\sin\alpha+y\cos\alpha$ and $C = a^2 B^2 + A^2 b^2 - a^2 b^2$.  Now the only thing I need to get is $h$ from (1). All other values are known, but I am not able to single that out.  Anyway if the above equations looks insane then please solve it yourself, your way. I could have drifted into some very complicated path.
117056
Consider the ellipse with equation:   $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$   How do we prove the product of distances from the foci to any tangent line is $b^2$?        $$F_1A \cdot F_2B = b^2$$
89348
Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.      Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.   OBS. $O_2$ is the zero matrix    Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $tr(A)=tr(B)=t$ and, from here, $A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2$  This is all I could get.  UPDATE  The matrices are from $M_2(\mathbb{R})$. Sorry for misleading you.
43327
I'm trying to see why this is the case. I have taken  the LHS and added $X\hat{\beta}$ and subtracted $X\hat{\beta}$.  From this I can get both terms on the right but I end up with a cross product term  $2(y-X\hat{\beta})^T(X\hat{\beta} - X\beta)$  that I can't prove equals to zero. I think there is some trick I'm missing, anybody remember it?