Find the shortest distance from the point $(1, 0)$ to the parabola $y^2 = 4x$.   As-  Let $D^2=(x-1)^2 + y^2$. Now reduce this to one variable by putting $y^2 = 4x$ to get  $D^2 = (x-1)^2+ 4x$. We are working with $D^2$ to  make the  calculations easier. Clearly $$\frac{d}{dx}D^2 = 2(x-1)+4 = 0 \implies x = -1$$ At $x = -1$, we have $\frac{d^2}{dx^2}D^2 = 2 &lt; 0$ which shows that $D^2$ is minimum. At this $x = -1$.  But this is not a point on parabola. Am i doing some mistake or we have to follow some other method to get the minima?
121353
Find a point on parabola $y^2 =4x$ which is at shortest distance from $(1,0)$. Answer given is $(0,0)$  But I am getting imaginary values. How do i get $(0,0)$
64533
I'm trying to find the minimal distance between the surfaces described by $z=x^2+y^2$ and $x+y-2z=8$. I would imagine there are several approaches including the use of Lagrange multipliers. I attempted to find the spot when their normal vectors are parallel ( since i believe if these vectors are not parallel there is a direction that will bring the distance lower). So if $U(x,y,z)=x^2+y^2$ and $V(x,y,z)=x+y-2z$ are functions describing these surfaces then their normal vectors are $(2x,2y,-1)$ and $(1,1-2)$ respectively. Then I want their cross product to be zero; this is at $(1/4,1/4,z)$. But when i solve for $z$ and I cannot simultaneously satisfy both equations. What gives?! Am I along the right line of thinking?
13337
I was solving a few problems from a textbook and I came across this one:     Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through   the point $P=(12,3)$   I could find the tangency points $$ but my solution was completely messy (I had four equations that I had to go back-and-forth with.  Is there an elegant/simple solution to that?  Tangency points:  $$P_1 = (0,3)$$ $$P_2 = \left( \frac{24}{5} , 3-\frac{24}{5} \right)$$
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My question is this:      $y^2 = \frac{x^5 - 1}{x-1}$ &amp; $x,y \in \mathbb{Z}$     Source: BdMO 2016 Regionals Set 1 Question 3.   I think this is the exact same question. Actually I didn't get the solution in the link. Please don't mark as duplicate.  I tried solving this same problem by writing code; there seems to be $6$ solution pairs to this equation. How do I get the $6$ solutions mathematically? I indeed need a full solution. Please help.
33337
Is there an elegant or easy way to derive the maximum distance from the origin to a point of the parabola created by projectile motion (assume initial position is the origin)? Other than differentiating with respect to $x$ and equate to zero the distance given by $\sqrt{x^{2}+y^{2}}$?
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I'm working through a classic Calculus book (Morris Kline), and one of the problems is:     Prove that the foot of the perpendicular from the focus to any tangent   of a parabola lies on the tangent to the vertex.   Basically, it's saying that if the parabola $y=x^2$ has a tangent line T at point $P$, and we draw a line $L$ perpendicular to $T$ that goes through the focus $F$, then $T$ meets $L$ at a point where $y=0$.  I managed to prove this by:   Calculating the equation of $T$ (based on slope of $y'$ and point $P$) Calculating the equation of $L$ (based on slope of $\frac{-1}{y'}$ and point $F$) Putting $y=0$ in both equations and solving for $x$ Observe that they both have the same $x$ at $y=0$ and therefore they meet on the $x$ axis.   HOWEVER, this seems like a very brute-force approach.  It's almost like I cheated or I simulated it on the computer with $1000$ points and determined that it is so.  I don't really understand WHY these $2$ lines must meet in this manner.  Is there any sort of geometric or intuitive proof?
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.
111342
I was given this question by a friend and after working tirelessly on it I have not come up with anything substantial. I was hoping someone in the community could provide a pointer or possibly a solution. The person who gave me the question says he has the answer to it, but I am more curious about the working.  Here is the question:     Three points $(B, C, D)$ lie on parabola  $y = x^2$ and a fourth point $A$ is placed such that $ABCD$ forms a square. Find the minimal area of $ABCD$.   So far my working out has involved a collection of distance/mid-point formula's and I haven't even reached the point where I differentiate to minimize the area.  Any pointers would be welcome.   Edit  Apparently the correct answer is $\sqrt{2}$ units squared.
700
I'm taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form  $$y = a x^2 + b x + c$$  My teacher gave me the formula:  $$x = -\frac{b}{2a}$$  as the $x$ coordinate of the vertex.  I asked her why, and she told me not to tell her how to do her job.  My smart friend mumbled something about it involving calculus, but I've always found him a rather odd fellow and I doubt I'd be able to understand a solution involving calculus, because I have no background in it.  If you use something you know from calculus, explain it to someone who has no background in it.  Because I sure don't.  Is there a purely algebraic or geometrical yet elegant derivation for the $x$ coordinate of a parabola of the above form?
112986
I am given an initial x and y position and initial velocity and I was asked to graph the trajectory in 1 second intervals.  This is what I have so far:  If $x_0 = 1, v_{0x} = 70, y_0 = 0, v_{0y} = 80, a_x = 0, a_y = -9.8$, and time will be 0,1,2,3... and so on.  Using these equations on every second you can find the plot will be a bell shaped with the highest point being ~ 325 m at about 600 seconds:   $$ x = x0 + (v_{0x})t + 1/2((a_x)t^2) $$ $$ y = y0 + (v_{0y})t + 1/2((a_y)t^2) $$  Usually in physics, we are taught in perfect condition with no air resistance. But what if there was air resistance? How would it affect this problem? How can i add it to my calculations and see what the difference is?
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This seems a bit obvious, but I'm not clear on one part. My teacher had told us $d(x^2) =2xdx$. However on asking why, we were told $d(x^2)= (x+dx)^2-x^2= 2xdx+(dx)^2$. We were told that we can ignore the $(dx)^2$ because its small. For obvious reason I feel this step is wrong, why can we ignore $(dx)^2$, is their any other way of getting the result?
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It's not too difficult to figure out that $x$ and $y$ can both be 1, and also $x$ can be 2 and $y$ can be 4 (and vice versa). But I can't rule out if there are other solutions. Does it have anything to do with inverse functions? Is there a way to see the solutions by graphing, or algebraically?
110913
An ant wants to get from (0,-2) to (0,2). However, there is a cone of radius 1 and height h at (0,0). What is the length of the shortest route?   Going around the cone uses $\frac{\pi}{3}+2\cdot \sqrt{3} ≈ 4.5$;   Going straight over takes $2+2\cdot \sqrt{h^2+1}$  Walking on side of cone might be shorter, but I cannot visualize it.
109848
If $ \frac {x^2}{a^2} + \frac {y^2}{b^2} = r^2$ is an ellipse, with the parameterization $x(θ)≔r(a \cos ⁡θ,b \sin⁡ θ ),$ I have to find the value of $θ$ giving the minimum distance from $P(p,q)$ (not on the ellipse) to the ellipse is given by a quartic in $t= \tan⁡( \frac {θ}{2}).$ A necessary condition for $x$ to be the closest point to $P$ is that $P-x$ is perpendicular to the tangent vector in $x ,$ i.e. $(P-x(θ) ). x' (θ)=0$  I can't handle the above condition to make an fuction (e.g. $f(θ)$) to find the minimum value by calculating the rerivative $f'(θ)=0$ for example. Then I have to prove that rational, non-zero values of $a, b, p, q$ can be found such that the quartic factorises as the product of two quadratics with rational coefficients. Any help? Thank you
76355
Given $x,y \in (-2,2)$ and $xy=-1$  Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$  My try:  Converting the function into single variable we get:  $$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$  $$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$  Using Differentiation we get:  $$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$  $$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$  $$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$  So the critical points are:  $x=0, x=\pm \sqrt{\frac{2}{3}}$  But $x \ne 0$ since $xy=-1$  $$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$  By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$  Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$  Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$  Is there any other approach?
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I have a simple parabola in the form $y = a + bx^2$. I would like to find the formula for a curve which is parallel to this curve by distance $c$. By parallel I mean that there is an equal distance along a line perpendicular to the tangent to my curve at all points.  I've established that the curve isn't in the form $y = a + c + dx^2$, whilst I can make this satisfy for $x=0$ and $x$ equal to one other number is isn't valid across the range.  Any help much appreciated.  Rob
83563
I have $(x_1, y_1), (x_2, y_2)$.  How do I find the point that's $d$ distance away from $(x_1, y_1)$ on a straight line to $(x_2, y_2)$?  I know I can get the length of the line with Pythagoras. I know if I drew a circle I could use the radius as distance and the point would be where the line and the circle intersect.  Could someone briefly explain each step to me please?  I don't understand