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0

you don't need to solve for $y$. you can implicitly differentiate $x^2/6 - y^2/8 = 1$ to get $$\dfrac{xdx}{3} - \dfrac{ydy}{4} = 0$$ substitute $x = 3, y = 2$ to find the tangent has slope $$dx - \frac{dy}{2} = 0$$ turn this into the tangent by replacing $dx, dy$ by $(x-3), (y-2)$ so that the equation of the tangent line at (3,2) is $$(x-3) - ...


0

If would compute the gradient of $x^2/6-y^2/8$ at the given point. If nonzero (which it is), it is perpendicular to the level set.


0

The best place to start, in my opinion, is by plotting this curve by hand. You'll soon realize that you're trying to take a line tangent to the curve $y=-x^{1/2}$, at the point $(16,-4).$ To find the slope of this curve at this point, we find the first derivative of the function $y=-x^{1/2}$: $$y'=-\frac{1}{2}x^{-3/2}$$ Evaluated at the point $(16,-4)$: ...


0

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 ...


0

You should first find dy/dx in terms of y and x. $y^2 = x \implies 2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y}$ Now, the slope of the tangent is $\frac{dy}{dx}$ evaluated at $y=-4$ which is $\frac{-1}{8}$, I think you can do the rest.


0

You write $y=m(x-16)-4$ (this gives parametrisation of all lines through $(-4,16)$). Replacing in $x-y^2$ you want get a polynomial in $x$ which should divide $(x-16)^2$. $x-y^2=(x-16)(-m^2(x-16)+8m+1)$, so $m=-\frac{1}{8}$, and the tangent line has equation $$y=-\frac{1}{8}(x-16)-4.$$


0

Use calculus: Differentiating gives $\frac{dy}{dx} = \frac1{2y}$. Thus the slope of the line is $\frac{1}{2y}$, which, at (16,-4) is equal to $-\frac{1}{8}$. Using the point-slope form of a line gives us: $y+4=-\frac18(x-16)$, where we have plugged in the point in question.


0

First figure out what kind of conic it is. Since the eccentricity of a parabola or circle is 1.0 (parabola) or 0.0 (circle), it's not one of those. It's either an ellipse or hyperbola, and since the eccentricity is less than 1.0, it's not a hyperbola. So it's an ellipse. Next use the focus and directrix to figure out the equation of the ellipse, given that ...


0

Use parametric form $P_0:(x_0=a\sec\alpha,y_0=b\tan\alpha)$ and $P_1:(x_1=a\sec\beta,y_1=b\tan\beta)$ Find the abscissa$(h)$ of the intersection of the tangents at $P_0,P_1$ to be $a\dfrac{\sin(\alpha-\beta)}{\sin\alpha-\sin\beta}$ Now apply the condition of the collinearity of $P_0,P_1,S$ to get $e =\pm\dfrac{\sin\alpha-\sin\beta}{\sin(\alpha-\beta)}$ ...


0

By a suitable stretching of the plane in the direction of the axis of one of the ellipses, you can turn this ellipse to a circle, while the other remains an ellipse. Now checking if the circle and the ellipse have a nonempty intersection is the same as checking if the center of the circle is inside the outward offset curve of the ellipse, at distance $r$. ...


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 ...


0

It would be given by $f(x, y, z) = h(x, y, z) = 0.$ You can also work out the kind of section, by determining how many lines in the cone are parallel to the plane: 0 for an ellipse, 1 for a parabola and 2 for a hyperbola.


0

let $(2a, 2a^2)$ is the point on $2y = x^2$ that has shortest point from $(-4,1).$ the slope of the tangent line at $(2a, 2a^2)$ is $2a$ this tangent must be perpendicular the line connecting $(-4,1)$ and $(2a, 2a^2).$ so the product of the slopes $$ 2a \dfrac{(2a^2 - 1)}{2a+4} = -1$$ solving this equation should give you $a = -1$ and the point $(-2, 2)$ on ...


0

Let $(x,\frac12x^2)$ be a point on the curve. The derivative at this point is $x$. And the slope of the line from $(-4,1)$ to $(x,\frac12x^2)$ is $\dfrac{\frac12x^2-1}{x+4}$. At the nearest point, these two slopes are perpendicular: $$x\cdot\dfrac{\frac12x^2-1}{x+4} = -1$$ This gives us $\frac12x^2-x = -x - 4$, which simplifies to $x^3=-8$, with the unique ...


0

Minimize the distance squared $F=(x+4)^2+(y-1)^2=(x+4)^2+(\frac{x^2}{2}-1)^2=\frac{x^4}{4}+8x+17$. $$\frac{dF}{dx}=x^3+8=0 --> x=-2$$. $$y=2^2/2=2$$


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

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

If what you mean by "touching" is "intersect", then this is the solution. Your circle is defined by its center $(x_c,y_c)$ and its radius $R$: $$(x-x_c)^2+(y-y_c)^2=R^2$$ Now, you want all three points $(1,0)$, $(4,4)$ and $(4,-4)$ to be on the circle, so you hav to solve the system: ...


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.


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 $a(t)$.


1

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


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 ...


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...


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

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 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.


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 $b^2=a^2(\varepsilon^2-1)$ ...


0

I found similar question in my textbook .hope you can follow on same lines


0

I claim that the point on the ellipsoid with the shortest distance to your plane will be such that the vector normal to the ellipsoid at that point will be parallel to the normal to the plane. The normal at (x, y, z) has the form (2x, 2y, 8z), and the normal to the plane is (1, 1, 1). Therefore, at the closest (and furthest) point on the ellipsoid, there is ...


0

Let $(r, s, t)$ be a point on the ellipsoid $x^2+y^2+4z^2=4$. Then, the signed distance from the point to the plane is $$d=\dfrac{r+s+t-6}{\sqrt{1^2 + 1^2 + 1^2}}.$$ Now we just need to minimize $d$ subject to $r^2+s^2+4t^2 - 4 =0$. Lagrange Multipliers should work here.


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: ...


0

The equation for a "wide ellipse" centered at $(h,k)$ is $$ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.$$ You're given that the ellipse is centered at the origin $(0,0)$, and $a=16$, so $$\frac{x^2}{16^2}+\frac{y^2}{b^2} = 1.$$ Now since $(8,6)$ is a point on the ellipse, setting $x=8$ and $y=6$ will give you an equation in one variable (namely $b$). You can ...


-1

In the Cartisian coordinates $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ describes an ellipse centered at the origin. You are given the value of $a$ and a point through which the ellipse passes. Can you find the value of $b$?


0

The eccentricity of a conic can be defined as the distance between the foci divided by the distance between the points of intersection of the conic with its major axis (its ends). In a circle, the foci are coincident at the center of the circle. Thus, $\epsilon=0$. In an ellipse, the foci are distinct and inside the ellipse and the ends are the points ...


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, ...


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) & = ...


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.


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 ...


0

general method will depend on how complex the function $f$ is. for simple polynomials you can find the intersection of a line and the graph of $y = f$ and require the repeating roots for the line to be a tangent. if the function is complex, you do the following steps: (a) you pick a point $(k, f(k)$ (b) get the tangent at this point $y - f(k) = (x - ...


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

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


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 ...


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 $|PA|=|Pm|$ ...


1

Your equation $y-y_1=a(x-x_1)^2$ comes from a horizontal directrix and a focus not on the directrix. So given those requirements, let the focus be the point $(x_1,y_1+d/2)$ and the directrix be the line $y=y_1-d/2$, where $d\ne 0$. This will put the vertex of the parabola at $(x_1,y_1)$ and the distance between the focus and the directrix will be $|d|$. ...


0

The parabola is defined by the locus of the points that are at equal distance from the focus and a line, called the directrice. Let's call $d$ the distance between the focus $F$ and the directrix $D$. Then $a=\dfrac{1}{2d}$


0

I found the equations a bit hard to understand. However, I can tell you how to find the other axis of an ellipse if you know one axis and one more point on the ellipse. An ellipse is obtained by stretching or expanding a circle along two perpendicular directions. Consider two perpendiculars trough the center of a circle. The distances from a point on the ...


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.


0

As said in the comments, this is not possible. The ellipse consists of a line which is neither a straight line (ruler) nor some arc of a circle (compass). It is however possible to draw an ellipse using two pins, a piece of rope and a pencil. Pin one end of the rope to the paper and pin the other end on the paper as well, in such a way that the rope is not ...


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 ...


0

It is because the principal symbol of the partial differential operator involved in the process. You can think a typical linear PDE as $$ Pu=0 $$ where $u$ is in some function space with possible boundary condition given, and $P$ is a linear differential operator like $\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial ...



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