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Problem 1
Find common tangent to the curve: $y+x^2=-4$ and $x^2+y^2=4$.
My idea:
Let $t1... y=ax+b$ is a tangent line to the first curve.
Let $t2... y=cx+d$ is a tangent line to the second curve.
But we are seracing for common tangent so that mean that $a=c$ and $b=d$.
By using implicit differentiation we can find $a$ and $b$. $$y+x^2=-4$$ $$y'+2x=0$$ $$y'=-2x$$ $$x^2+y^2=4$$ $$2x+2yy'=0$$ $$x+yy'=0$$ $$y'=\frac{-x}{y}$$ $$-2x=\frac{-x}{y} \to y=\frac{1}{2},$$$x$ is any real number Problem 2
For which value of the coefficients $a$, $b$ and $c$ $\in$ R is the x-coordinate axis tangent to the curve?
$y=ax^2+bx+c$
My idea:
Curve and the tangent line must have just one common point. Our curve is a parabola and the number of dots on x axis is a number of solution of quadratic equation. So we need $a,b,c$ to be equal to $b^2-4ac=0$

Problem 3
Find a line that is tangent to the curve $y = x^4 - 2x^3 - 3x^2 + 5x + 6$ in at least two points .
My idea:
Let that tangent be $t=ax+b$. If we that line be a tangent line at two point to the curve equation $y(x)-t(x)=0$ must have at least two (not equal) solution. $$x^4 - 2x^3 - 3x^2 + 5x + 6-(ax+b)=0$$ $$x^4 - 2x^3 - 3x^2 + 5x + 6=ax+b$$ $$a=5 , b=6$$ $$x^4 - 2x^3 - 3x^2=0$$ $$x_1=0, x_2=-1, x_3=3$$

Problem 4 I am trying to find the number of tangents to a curve that all pass through the origin. The curve's equation is $y=x^3+x^2−22x+20.$ I also need to find the equation of said tangents. My work: Let's use formula for tangent line: $$y-y_0=y'(x_0)(x-x_0)$$ We know that $x=0$ and $y=0$.
Let's find $y'(x_0)$: $y'(x_0)=3x_0^2+2x_0-22$
Plugin what we know: $$-y_0=(3x_0^2+2x_0-22)*-x_0$$ We aslo know that $y_0=x_0^3+x_0^2-22x_0+20$ $$-x_0^3-x_0^2+22x_0-20=-3x_0^3-2x_0^2+22x_0$$ $$2x_0^3+x_0^2-20=0$$ Only real solution is 2. $x_0=2$ and $y_0=-12$. Let tangent line be $y=kx+l$.
$k=y'(2)=-6$
$-12=-6*2+l\to l=0$ But we want l to be zero. Our solution is $y=-6x$.

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  • $\begingroup$ What's your question? $\endgroup$
    – bubba
    Apr 3 '16 at 9:00
  • $\begingroup$ Is my work on problem 2 and 3 ok? $\endgroup$
    – josf
    Apr 3 '16 at 9:08
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For problem 1, note that the requirement is that two points(not necessarily the same) must share a tangent line(which means both $a=c$ and $b=d$). That they might not be the same point means you cannot assign them the same x and y values. So we have to simultaneously solve: $$-2x_1=x_2/y_2$$ And: $$b=d$$ To find expressions for b and d, we can use: $$y=y_k'(x-x_k)-y_k$$ That is, $b=x_1y_1'-y_1$ and $d=x_2y_2'-y_2$.

You have to simultaneously solve: $$-2t=-u/\sqrt{4-u^2}$$ And: $$-t^2+4=-u^2/\sqrt{4-u^2}-\sqrt{4-u^2}$$

Wolfram is giving me a double surd solution to these equations. This seems pretty nasty to solve. Have you got the questions right?

Your answer to problem 2 is correct and well reasoned. =]

For problem 3, you need to find a line $y=ax+b$ such that $P(x)-y(x)=(x-\alpha)^2(x-\beta)^2$ $$x^4-2x^3-3^2+(5-a)x+(6-b)\equiv(x-\alpha)^2(x-\beta)^2$$

Expanding the right hand side: $$x^4-2x^3-3^2+(5-a)x+(6-b)\equiv x^4-2(\alpha+\beta)x^3+(\alpha^2+\beta^2+4\alpha\beta)x^2-2(\alpha^2\beta+\alpha\beta^2)x+\alpha^2\beta^2$$

We can find $\alpha$ and $\beta$ by solving: $$-2(\alpha+\beta)=-2$$ And: $$\alpha^2+\beta^2+4\alpha\beta=-3$$ Giving us $\alpha=1$ and $\beta=-2$.

From there: $$x^4-2x^3-3x^2+(5-a)x+(6-b)\equiv x^4-2x^3-3x^2+4x+4$$

That is, $5-a=4\to a=1$ and $6-b=4\to x=2$. Our tangent is $y=x+2$

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  • $\begingroup$ Thanks, I update my question to problem 4 so please check is work ok? $\endgroup$
    – josf
    Apr 3 '16 at 10:24
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    $\begingroup$ Looks good to me. =] $\endgroup$ Apr 3 '16 at 10:30
  • $\begingroup$ What about equal example but my function is given implicitly. How can I work out that? For example $x^2y+y=2$ $\endgroup$
    – josf
    Apr 3 '16 at 10:34
  • $\begingroup$ Just solve for y. If you can't solve for y in a case like that, it may just be too difficult. There might be a general solution in multivariable calculus but I'm not sure. You can usually find the derivative of y since, if $f(x,y)=0$, $y'=-f_x(x,y)/f_y(x,y)$. For now, this seems complicated enough. $\endgroup$ Apr 3 '16 at 11:05
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For problem 1 :

After you have $y'=-2x,y'=-x/y$, you have $-2x=-x/y\Rightarrow y=1/2$, but this is wrong because the $x$ on LHS is not necessarily the same as the $x$ on RHS.

Let $(p,q)$ be the tangent point on $y+x^2=-4$, and let $(s,t)$ be the tangent point on $x^2+y^2=4$.

Then, the equation of the common tangent can be written as $$y=-2px+2p^2+q=-\frac stx+\frac{s^2}{t}+t$$ Now, solve the following system : $$q+p^2=-4\tag1$$ $$s^2+t^2=4\tag2$$ $$-2p=-\frac st\tag3$$ $$2p^2+q=\frac{s^2}{t}+t\tag4$$ From $(1)(2)(3)$, representing $p^2,q,s^2$ by $t$,

$$p^2=\frac{-t^2+4}{4t^2},\quad q=\frac{-15t^2-4}{4t^2},\quad s^2=4-t^2$$ From $(4)$, $$2\times\frac{-t^2+4}{4t^2}+\frac{-15t^2-4}{4t^2}=\frac{4-t^2}{t}+t,$$ i.e. $$17t^2+16t-4=0\quad\Rightarrow\quad t=\frac{-8\pm 2\sqrt{33}}{17}$$ from which it follows that $$(t,p^2,q,s^2)=\left(\frac{-8\pm 2\sqrt{33}}{17},12\pm 2\sqrt{33},-16\mp 2\sqrt{33},\frac{960\pm 32\sqrt{33}}{289}\right)$$ Hence, the answer is $$\color{red}{y=2\sqrt{12+2\sqrt{33}}\ x+8+2\sqrt{33}}$$ $$\color{red}{y=-2\sqrt{12+2\sqrt{33}}\ x+8+2\sqrt{33}}$$ $$\color{red}{y=2\sqrt{12-2\sqrt{33}}\ x+8- 2\sqrt{33}}$$ $$\color{red}{y=-2\sqrt{12-2\sqrt{33}}\ x+8-2\sqrt{33}}$$


For problem 2 :

You are correct except that you have to have $a\not=0$.


For problem 3 :

You are not correct. We have to find $a,b,c,d$ such that $$x^4-2x^3-3x^2+5x+6-(ax+b)=(x-c)^2(x-d)^2,$$ i.e. $$x^4-2x^3-3x^2+(5-a)x+6-b$$$$=x^4+(-2c-2d)x^3+(c^2+4cd+d^2)x^2+(-2c^2d-2cd^2)x+c^2d^2$$ Now solve the following system : $$-2=-2c-2d\tag5$$ $$-3=c^2+4cd+d^2\tag6$$ $$5-a=-2c^2d-2cd^2\tag7$$ $$6-b=c^2d^2\tag8$$

From $(5)(6)$, $$c+d=1,cd=-2\quad\Rightarrow \quad (c,d)=(2,-1),(-1,2)$$ from which $a=1,b=2$ follow, and so the answer is $\color{red}{y=x+2}$.

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  • $\begingroup$ With Gröbner bases? ;o) $\endgroup$
    – Bernard
    Apr 3 '16 at 10:17
  • $\begingroup$ @Bernard: I added some details. Take a look. $\endgroup$
    – mathlove
    Apr 3 '16 at 11:30
  • $\begingroup$ I was only kidding. Might as well have proposed the resultant of $p,p'$… More seriously, the geometric approach is more telling, in my opinion. $\endgroup$
    – Bernard
    Apr 3 '16 at 11:42
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Another solution for problem 3:

Let solve the more general problem:

Let $p(x)$ be a polynomial of degree $n\ge 4$. Find bitangents to the curve $y=p(x)$.

Let $y-p(x_0)=p'(x_0)(x-x_0)$ be the equation of the tangent at the point $(x_0,p(x_0))$. It will be a bitangent if the abscissae equation for the intersection points of this line with the curve has another double root.

We'll write this equation using the Taylor's expansion of $p(x)$ at $x=x_0$: $$p(x)=p(x_0)+p'(x_0)(x-x_0)+\frac{p''(x_0)}{2}(x-x_0)^2+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^n.$$ The equation for the intersection points is thus $$\frac{p''(x_0)}{2}(x-x_0)^2+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^n$$ and the other intersection points abscissae are roots of $$\frac{p''(x_0)}{2}+\frac{p'''(x_0)}{6}(x-x_0)+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^{n-2}$$ So the abscissae of the other points of contact with the curve are the multiple roots of this equation which are different from $x_0$.

In the present case, as $n=4$, you get a quadratic equation, which must have a double root in $X=x-x_0$: $$6x_0^2-6x_0-1+(4x_0-3)X+X^2=0,$$ whence the condition $$(4x_0-3)^2-4(6x_0^2-6x_0-1)=-8x_0^2+13=0.$$

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Problem$\#1:$

The equation of tangent of $$x^2+y+4=0$$ at $(t,-(4+t^2)),$ $$x(t)+\dfrac{y+\{-(4+t^2)\}}2+4=0\iff2t x+y+4-t^2=0$$

Now this will be a tangent of $$x^2+y^2=4$$ if the perpendicular distance of tangent to center of circle $=$ radius of circle

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