Tangents are drawn from the point ($\alpha$,$\beta$) to the hyperbola  $3x^2 - 2y^2 = 6 $ and are inclined at angles A and B to the x -axis. If $\tan A. \tan B = 2$, prove that $\beta^2$=2$\alpha^2$ - 7.  I tried the following concept,The points on Hyperbola from where tangent are drawn are $(\sqrt2*(\sec A),\sqrt3*(\tan A))$ &amp;  $(\sqrt2*(\sec B),\sqrt3*(\tan B))$   The two tangents are $\frac{\sec A}{\sqrt2}$-$\frac{\tan A}{\sqrt3}$=1 and  $\frac{\sec B}{\sqrt2}$-$\frac{\tan B}{\sqrt3}=1$, I tried entering ($\alpha$,$\beta$) and using condition $\tan A. \tan B = 2$ but not able to get the result $\beta^2$=2$\alpha^2$ - 7.
123066
Tangents are drawn from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ and are inclined at angles $\theta$ and $\phi$ to the $x-$axis.If $\tan\theta.\tan\phi=2,$ prove that $\beta^2=2\alpha^2-7$    The equation of the pair of tangents from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ is given by $(3x\alpha-2y\beta-6)^2=(3x^2-2y^2-6)(3\alpha^2-2\beta^2-6)$ When i simplify this expression,i get $(18+6\beta^2)x^2+(6\alpha^2-12)y^2-12xy\alpha\beta+24y\beta-36x\alpha+18\alpha^2-12\beta^2=0$ This is an equation of pair of lines and the angle between pair of lines $ax^2+2hxy+by^2=0$ is given by the formula $\frac{2\sqrt{h^2-ab}}{a+b}$ So $\tan(\theta-\phi)=\frac{2\sqrt{36\alpha^2\beta^2-(18+6\beta^2)(6\alpha^2-12)}}{18+6\beta^2+6\alpha^2-12}$ $\frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}$ $\frac{\tan\theta-\tan\phi}{3}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}$ I am stuck here.I do not know,how to prove it further.
87980
I am trying to find the equation of a parabola with the two tangent equations to two points.  I found this post () but it did not really help me so I tried to do the math on my own.  First:   I have two points whose coordinates I know, I have two lines passing through these points whose slope I know because of their angles of inclination.   Let's say that these lines are tangents to a parabola in these two points.  How can I find the equation of this parabola given the two tangent equations and the two points ?  Here is a picture of my calculations:    According to the picture I drew I am suppose to get the following equation:  $f(x) =\dfrac{\tan(\alpha)-\tan(\beta)}{2 \cdot x_1}\cdot x^2 + \tan(\beta)\cdot x + y_2$  When I plot $f(x) =\dfrac{\tan(60)-\tan(30)}{-2 \cdot 10}\cdot x^2 + \tan(30)\cdot x + 10$ I get the following graph:    As you can see I do not get the graphic drawn in the first picture, and if I set the window to $x\in\left[x_1,0\right]$ and $y\in\left[0,y_2\right]$ I get an empty window:    What am I doing wrong ?  I hope my explanations are good enough. Please tell me if I was not clear so that I can modify my post for you and others.
29025
Show that if $x, y,$ and $z$ are consecutive terms of an arithmetic sequence, and $\tan y$ is defined, then  $$\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y. $$  I'm not sure what trig identities I would use and how to use them.  Could I get some help?  Thanks.
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?
71195
Triangle $ABC$, $AB=4$, $BC=15$, $AC=13$. Two sides are tangents to the respective Parabolas. We have to find the area shaded. My approach-  I tried finding the area of the quadratures(Archimedes) formed but it doesn't help as- I have to find the area between the 2 intersecting Parabolas also which I can't find.
120045
I'm not sure where to begin on this problem - do I plug in for a and solve for z?  I was also given a hint:  Let z be a point on the line we're trying to describe. We have good tools in complex numbers for collinearity and perpendicularity. Which would be useful here.  Here is the problem:  Let a and b be two complex numbers on the unit circle, i.e. $|a| = |b| = 1.$  (a) Show that the equation of the tangent to the unit circle at a is given by $ z + a^2 \overline{z} = 2a.$    (b) Show that the intersection of the tangents to the unit circle at a and b is $\frac{2ab}{a + b}.$
47039
Just doing this for revision, seems much harder than it should be, should I use $x=\tan u$ ? Any help appreciated.
40398
I am having some trouble understanding Math.tan() and Math.atan() and Math.atan2().  I have basic knowledge of trigonmetry but the usage of SIN, COS, and TAN etc for game development is very new to me.  I am reading on some tutorials and I see that by using tangent we can get the angle in which one object needs to be rotated by how much to face another object for example my mouse. So why do we still need to use atan or atan2?
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
16150
How prove  this inequality $$\tan{(\sin{x})}&gt;\sin{(\tan{x})},0&lt;x&lt;\dfrac{\pi}{2}$$  this  PDF give  a ugly methods :   I post this solution:let     $$f(x)=\tan{(\sin{x})}-\sin{(\tan{x})}$$   then we have   $$f'(x)=sec^2{(\sin{x})}\cos{x}-\cos{(\tan{x})}\sec^2{x}=\dfrac{\cos^3{x}- \cos{(\tan{x})}\cos^2{(\sin{x})}}{\cos^2{(\sin{x})}\cos^2{x}}$$   case 1:$0&lt;x&lt;\arctan{\dfrac{\pi}{2}}$, then we have    $$0&lt;\tan{x}&lt;\dfrac{\pi}{2},0&lt;\sin{x}&lt;\dfrac{\pi}{2}$$   so Use AM-GM inequality we have   $$\sqrt[3]{\cos{(\tan{x})}\cos^2{(\sin{x})}}\le\dfrac{1}{3}[\cos{(\tan{x})}+2\cos{(\sin{x})}]\le\cos{\dfrac{\tan{x}+2\sin{x}}{3}}$$   use   $$\tan{x}+2\sin{x}&gt;3x$$   so   $$f'(x)&gt;0\Longrightarrow f(x)&gt;0$$   case2: $\arctan{\dfrac{\pi}{2}}\le x\le\dfrac{\pi}{2}$, so   $$\sin{(\arctan{\dfrac{\pi}{2}})}&lt;\sin{x}&lt;1$$   since   $$\sin{(\arctan{\dfrac{\pi}{2}})}=\dfrac{\pi}{\sqrt{4+\pi^2}}&gt;\dfrac{\pi}{4}$$   $$\Longrightarrow \dfrac{\pi}{4}&lt;\sin{x}&lt;1$$,so   $$1&lt;\tan{(\sin{x})}&lt;\tan{1},$$,so   $$f(x)&gt;0$$   I think this inequality have other  simple methods.Thank you  It is said can use integral inequality,But I can't   In fact,we have $$\tan{(\sin{x})}-\sin{(\tan{x})}=\dfrac{1}{30}x^7+o(x^7),x\to 0$$ can see:
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When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of the squares of the semimajor and semiminor axes, i.e. $c^2=|a^2-b^2|$ for the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.) The picture argument goes as follows: just draw the obvious isosceles triangle from the two foci to one of the two points at the end of the minor axis. We see two right triangles; in each, one leg is the focal length $c$ and the other leg is the semiminor axis, while the hypotenuse is the semimajor axis. The result follows from the Pythagorean theorem applied to this right triangle.  But when it comes to the Pythagorean relation for hyperbolas, I can't think of an equally convincing picture to draw. Instead I have to give a rather inelegant algebraic argument for the relation $c^2=a^2+b^2$ (for the hyperbola defined by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with focal length $c$), directly from the definition of the hyperbola. The messy details are below.*  Now, obviously there is an asymmetry between the argument for the ellipse and the argument for the hyperbola: the ellipse argument is almost trivial and doesn't involve working straight from the locus definition, but the hyperbola argument is tedious and does require working straight from the definition. So, my question:     Is there an equally persuasive picture I could draw to convince high   school students of the Pythagorean relation for hyperbolas -- without having to get technical with points at infinity, say?   The natural triangle to draw is the right triangle with legs from the center to a vertex and from that vertex to an asymptote, which has legs of length $a$ and $b$; but then the problem becomes why the length along the asymptote should be equal to $c$. I doubt there is a purely (Euclidean) geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some sort of limiting procedure will be required.    [*] The tedious algebraic details:     The foci lie at $(\pm c,0)$.   Say $a&gt;b$. Then hyperbola is the locus of points $(x,y)$ such that       $$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$      Assume for simplicity the expression between the absolute value bars   is positive. (The algebra is even more tedious if we don't make such a   simplifying assumption.) Then moving the right root to the other side   and squaring gives      $$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$      Simplifying gives      $$xc-a^2=a\sqrt{(x-c)^2+y^2}$$      and upon squaring again we have      $$x^2c^2-2xca^2+a^4=a^2(x^2-2xc+c^2+y^2)=a^2x^2-2xca^2+c^2a^2+a^2y^2$$      which reduces to      $$x^2(c^2-a^2)-y^2a^2=a^2(c^2-a^2)$$      Defining $b^2:=c^2-a^2$ brings the equation into the form   $x^2/a^2+y^2/b^2=1$, and the Pythagorean relation follows from the   definition of $b$.
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I have a hyperbola in the first quadrant opening to the right given with the equation $$x=\sqrt{19.8y^2-6.9y+0.7}$$ and I have to find the tangent line from the point $(0, 0.08)$ to the hyperbola (the tangent line with positive slope, so above the hyperbola if it's opening to the right).     The problem is I don't know what the tangent point is on the hyperbola because then I would just be able to find the slope and use $y=mx+b$ to find the equation of the tangent line.  My idea is since there are many lines that intersect the hyperbola from that point I have to maximize the slope between the point and the hyperbola and then I should be able to find the tangent line between $(0, 0.08)$ and the hyperbola. But I don't know how to find the maximum slope between the line that goes through the point and the hyperbola?
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.
63583
My (very simple) question to a friend was how do I prove the following using basic trig principles:  $\tan75^\circ = 2 + \sqrt{3}$  He gave this proof (via a text message!)  $1. \tan75^\circ$   $2. = \tan(60^\circ + (30/2)^\circ)$  $3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ) $  $4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$    Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ?    EDIT  Using the answers given I'm almost there:   $\tan75^\circ$    $\tan(45^\circ + 30^\circ)$   $\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$   $(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$    $\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$   $\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$   multiply throughout by $(\sqrt{3}) + 1)$   Another alternative approach:   $\tan75^\circ$    $\tan(45^\circ + 30^\circ)$   $\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$   $\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$   at point 6 in above alternative
107848
How to find this ?   I'm stuck at $1$ and how to $1$ to change the tangent associated with a,b or c.  I meant this ,everybody.  $\frac{tan(a)+tan(b)+tan(y)−tan(a)tan(b)tan(y)}{1−tan(a)tan(b)−tan(a)tan(y)−tan(b)tan(y)}$
84191
Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.  Attempt at a solution:  $[sin y = 3(cos x cos y - sin x sin y) sin x] 1/cos y$  $tan y = (3 cos x - 3 sin x tan y) sin x$  $tan y + 3 sin^2 x + tan y = 3 sin x cos x$  $tan y = (3 sin x cos x) / (1 + 3 sin x)$     I'm stuck here...
92237
Hi all, I am interested to find elementary proof of tangent half angle formula.  My solutions are the following:  Triangle $AOB$ is such that $|AB|=1$ and $\angle AOB=\theta$. We then extend $OB$ to $P$ and $Q$ such that $|OP|=|OQ|=1$. Thus we will have two isosceles triangles: $AOP$ and $AOQ$.  From the picture, $\tan{\left(\frac{\theta}{2}\right)}=\frac{AB}{BP}=\frac{\sin{\left(\theta\right)}}{1+\cos{\left(\theta\right)}}\ \ \ =\frac{BQ}{AB}=\frac{1-\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}$  Could You guys please check my solution. I am also wondering if there are other elementary solutions, please share, thanks!