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5

The general form of a parabola/quadratic is: $$y=ax^2+bx+c$$ You should be able to construct a pair of simultaneous equations using the two points and solve for $a,b,c$. Another method is to consider: $$y=\lambda(x-\alpha)(x-\beta)$$ Where $\alpha , \beta$ are the roots of the parabola.


4

Let $A$ be the center of $C$, and $r$ be its radius. For any point $P$, let $|Pl|$ be the perpendicular distance from $P$ to $l$. There are two ways a point $P$ can be on your locus. Either (1) the distance $|PA|=|Pl|+r$, or (2) $|PA|=|Pl|-r$. These two cases can be treated one by one. For case (1), the condition $|PA|=|Pl|+r$ is the same as $|PA|=|Pm|$ ...


3

Hint: $\vec{v(t)} = \vec{r'(t)}, v(t) = ||\vec{r'(t)}||, \vec{a(t)} = \vec{r''(t)}, a(t)=||\vec{r''(t)}||$. Thus: $\vec{v(t)} = (2\cos t, -3\sin t), v(t) = \sqrt{4\cos^2t+9\sin^2t}$,etc...


3

The notation $\vec{r}(t)$ is very common to denote the position of a particle as a function of time. For example, $\vec{r}(t) = (\cos t, \sin t)$ with $t\in [0,2\pi]$ describes a particle moving counter-clockwise around the unit circle (you can check this with a parametric graph, or noting that the magnitude of $\vec{r}(t)$ (which is the particle's distance ...


2

We first take the equation for an eclipse centered at $(h,k)$: $$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.$$ You are given that the ellipse is centered at the $(0,0)$, $a=16$, and $(8,6)$ is a point on the ellipse. Hence $$\frac{8^2}{16^2}+\frac{6^2}{b^2} = 1.$$ Now from this, we can solve for $b$ to get the final equation: ...


2

Given a parametric curve $(x(t), y(t))$, the (signed) curvature is given by \begin{equation*} \kappa(t) = \frac{x' y'' - y' x''}{((x')^2 + (y')^2)^{3/2}}. \end{equation*} The ellipse $9(x-1)^2 + y^2 = 9$ can be rewritten in standard form as $(x-1)^2 + \frac{y^{2}}{9} = 1$. Then parametrize it as \begin{align*} x(t) & = 1 + \cos{t}\\ y(t) & = ...


2

Did you try a less clever method? Put $x=\alpha x' + \beta y'$ and $y=-\beta x' + \alpha y'$ Then the equation $5x^2 + 5y^2 - 6xy - 8 = 0$ becomes $(5(\alpha^2+\beta^2)+6\alpha\beta)x'^2+(5(\alpha^2+\beta^2)-6\alpha\beta)y'^2-6x'y'(\alpha^2-\beta^2)-8=0$ If you want to remove any cross product, then $\alpha^2-\beta^2=0$ EDITED after abel comment ...


2

Hint I did what you tell $(x=p+q , y=q-p)$ and the raw result I obtained for the expression is $$4 p^2+32 p+8 q^2-32 q+92=0$$ I let you the task of grouping but it seems that you made some mistakes. I am sure that you can take from here and fix the problem.


2

the tangent to the parabola at $(4, 4)$ has slope $1/2$ so the radius has slope $-2.$ let the center of the circle touching the parabola $y^2 = 4x$ at $(4,4)$ be $x = 4 + t, y = 4 - 2t$. now equating the radius $$5t^2 = (3+t)^2 + (4-2t)^2$$ you can find $t$ which will give you the center and the radius of the circle.


2

As $y=3$ contains the foci, it also contains the major axis If the equation is $\dfrac{(x-\alpha)^2}{a^2}-\dfrac{(y-\beta)^2}{b^2}=1$ As the center is the midpoint of the foci, we have $2\alpha=-1+3,2\beta=3+3$ Now the coordinates of the foci are $(\alpha\pm a\varepsilon,\beta)$ So, $1+2a=3,1-2a=-1\implies a=1$ We know $b^2=a^2(\varepsilon^2-1)$ ...


2

Equation of a standard downward facing parabola is $(x-x_1)^2=-4a(y-y_1),$ and in that parabola the perpendicular distance from the focus to the directrix is $2a$, In your case, $(x_1,y_1)=(-3,23)$ and $4a=16$ which is double of $2a=8.$


2

$(y-23) = -\frac{1}{16}(x+3)^2$ is a parabola that is congruent to $y=-\frac{1}{16}x^2$, which is much easier to analyse. The focus of the latter parabola is at $F(0,-p)$, and the closest point on the directix from an arbitrary point $P(x,-\frac{1}{16}x^2)$ is $D(x,p)$. Of course, the vertex is at V(0, 0). Note that the distance from $P$ to $D$ is ...


2

If $F_2$ is any point such that $F_1P_1 + F_2P_1 = F_1P_2 + F_2P_2$, then there exists an ellipse through $P_1$ and $P_2$ with foci $F_1$ and $F_2$. $F_2$ satisfies $F_2P_1 -F_2P_2 =$ constant, and the locus of all such points is a circle. So there is no unique solution.


1

For sufficiently regular functions such as polynomial or rational functions, you can take a straight line with a variable slope $t$ passing though $(a,b)$ and determine its points of intersection with the curve $y=f(x)$. You have a tangent line if the resulting equation has a double root. In the example you give, you obtain the equation $t^2x^2=4a(x-a)\iff ...


1

The co-ordinates of the focii are $(h\pm ae)$, so your $c$ will be $ae$ rather than $\frac{e}{a}$ and so your $a$ will be $\sqrt2$ and your $b$ will be $\sqrt\frac{3}{2}$. So the equation we have in the rotated system is:$$\frac{(x-\frac{1}{\sqrt2})^2}{2}+\frac{2y^2}{3}=1$$ Now all that remains is to somehow rotate this back into the original system.


1

You can use the general equation of a line through the point(16, -4). It will intersect the parabola in $(16,-4)$ and another point. The tangent will correspond to the case when this other point is $(16,-4)$ again, which will give a condition on $m$ for this to happen. A line with slope $m$ passing through $(16,-4)$ has equation $\,\,y+4=m(x-16)$. To find ...


1

You can construct an arbitrary number of points on an ellipse using e.g. Pascal's theorem (which doesn't even require compasses). If you want the ellipse as a curve, though, not as a set of points, then this is strictly speaking impossible since both ruler and compasses can only draw curves of constant curvature, which the ellipse is not. You can, of course, ...


1

here is a compass and rule construction of of an ellipse with major axis $a$ and foci at $F_1 = (-ae, 0)$ and $F_2 = (ae, 0)$ here are the construction steps: (a) pick two points $F_1, F_2.$ draw the line $F_1F_2$ (b) draw a circle with center $F_1$ and radius big enough to contain $F_2$ within it. let the circle cut $F_1F_2$ extended at $A$ so that $F_2$ ...


1

Well, you have $TSR$ and $TR'S'$ similar, hence $\dfrac{RS}{R'S'}=\dfrac{TS}{TS'}$ You have also, as you stated, $\dfrac{TS}{TS'}=\dfrac{PS}{PS'}$ Then you have $\dfrac{RS}{R'S'}=\dfrac{PS}{PS'}$ That is also $\dfrac{R'S'}{PS'}=\dfrac{RS}{PS}$. Can you finish from here?


1

The center lies at the intersection of the perpendicular to the tangent and the bisector of the two known points. $$2(x-4)+(y-4)=0$$ $$(4-1)(x-\frac{4+1}2)+(4-0)(y-\frac{4+0}2)=0.$$ Hence $$x=\frac{13}2,y=-1.$$


1

A non standard solution Take a second point on the parabola very close to $(4,4)$: $(4+\epsilon,4+2\epsilon)$ agrees to the first order (plugging in the equation of the parabola, $16+8\epsilon+\epsilon^2\approx16+8\epsilon$). Then solve the circle by the three points $$\begin{align} (x-4)^2&+(y-4)^2&=r^2,\\ ...


1

I read the question again. I realize now the implicit mistake. Generically, provided $F$ is reasonably differentiable, any equation of the form $F(x,y) = 0$ gives a curve in the plane. If you change the exponent to say $Ax^3+\cdots +C=0$ then you would still have a curve in the plane. I suppose the question you are really asking: is such a curve interesting ...


1

Let we suppose that $E_1$ is an ellipse with equation $f(x,y)=\frac{x^2}{a}+\frac{y^2}{b}-1=0$ and $E_2$ is another ellipse. To check if $E_1$ and $E_2$ intersect, it is sufficient to check if $f(x,y)$ takes only positive values on $\partial E_2$. So we can take a parametrization of $\partial E_2$ and compute the stationary points for the quadratic function ...


1

(just comments transfered to answer) s $ \phi$ cannot be represented by an angle. if only because it has a range $ − \infty < \phi < \infty $ , while angles have a range $ − \pi < x < \pi $ ) $ \phi$ can be represented by an area see en.wikipedia.org/wiki/Hyperbolic_function and math.stackexchange.com/a/451372/88985 . $ \phi$ is twice the ...


1

Given $$Ax^2+Bx+Cy^2+Dy+Exy+F=0,$$ Then $$tan(2\theta)=\frac{E}{C-A},$$ for $$x=\bar{x}cos(\theta)+\bar{y}sin(\theta),\quad y=-\bar{x}sin(\theta)+\bar{y}cos(\theta).$$


1

define new variables $$x = x_1\cos t - y_1 \sin t , y = x_1 \sin t + y_1 \cos t$$ substituting in $5(x^2+y^2) - 6xy - 8= 0$ you get $$5(x_1^2 + y_1^2) - 6(x_1\cos t - y_1\sin t)(x_1\sin t + y_1 \cos t) - 8 = 0$$ which simplifies to $$x_1^2(5 - 6 \sin t \cos t) + y_1^2(5 +6\sin t \cos t)- 6x_1y_1(\cos^2 t - \sin ^2 t) - 8 = 0$$ you can eliminate the $x_1y_1$ ...


1

It's not extremely difficult, just take care not to have a square root everywhere: $$ \sqrt{(x-1)^2+y^2}+\sqrt{(x-3)^2+y^2}=6 \\ \sqrt{(x-1)^2+y^2} = 6 - \sqrt{(x-3)^2+y^2} \\ (x-1)^2+y^2 = 36 + (x-3)^2+y^2 - 12\sqrt{(x-3)^2+y^2} \\ x^2-2x+1+y^2-36-(x^2-6x+9)-y^2 = -12\sqrt{(x-3)^2+y^2} \\ -12\sqrt{(x-3)^2+y^2} = 4x-44 \\ 9\left[x^2-6x+9+y^2\right] = ...


1

The focus-directrix definition of a conic section (which seems to be the definition that you're referring to) is as follows: Given a line $L$, a point $P$, and a real number $\varepsilon>0$, a conic section is the locus of all points such that the distance to $P$ is $\varepsilon$ times the perpendicular distance to $L$. $P$ is referred to as the focus, ...


1

$$\vec v(t) = \dfrac{\text{d} \vec r}{\text{d} t} = (2 \cos t, \, -3 \sin t)$$ $$v(t) = ||\vec v(t) || = \sqrt{2^2 \cos^2 t + (-3)^2 \sin^2 t}$$ $$\vec a(t) = \dfrac{\text{d}^2 \vec r}{\text{d} t^2} = \dfrac{\text{d} \vec v}{\text{d} t} = (-2 \sin t, \, -3 \cos t)$$ $$a(t) = ||\vec a(t) || = \sqrt{(-2)^2 \sin^2 t + (-3)^2 \cos^2 t}$$


1

$\vec r(t) = (2\sin t, 3\cos t)$ HINT: $\vec v(t) = \dfrac{d \vec r(t)}{dt}$$= \dfrac {d}{dt}$ $(2\sin t, 3\cos t)$ From this, you can find $v(t)$. Using Formula : If $\vec r=(a,b)$ then $|\vec r|=r=\sqrt {a^2 + b^2}$ And, $\vec a(t)=\dfrac{dv(t)}{dt}$ .... and you will get $a(t)$.



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