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Problem 1
On the curve $y=\frac{1}{1+x^2}$ find a point in which tangent line is parallel to the horizontal axis.
My idea:
Let's find $y'$. $$y'=\frac{-2x}{(1+x^2)^2}$$ If we want a tangent line to be parallel to x-axis it must look like this $t...y=kx+l$ such that $k=0$. Let's find for which $x$ we have that situation. $$y'=\frac{-2x}{(1+x^2)^2}=0$$ $$x=0\to y=1$$ $$(x,y)=(0,1)$$ Problem 2 Determine the angle of the curve $y^2 - 3x^2 + x + 1 = 0$ and $xy^2 - 1 = 0$ intersect in the first quadrant
My work: $y^2 - 3x^2 + x + 1 =xy^2 - 1$ One solution for $x>0$ and $y>0$ is $x=y=1$.
Let's now find derivative of easch function at that point. $$2xyy'=-y^2$$ $$y'=\frac{-y^2}{2xy}=\frac{-1}{2}$$ $$2yy'=6x-1$$ $$y'=\frac{6x-1}{2y}=\frac{5}{2}$$ And the angle is $\alpha =arctg|\frac{k_1-k_2}{1+k_1*k_2}|$ where $k_1=\frac{-1}{2}$ and $k_2=\frac{5}{2}$
Problem 3
Find the angle of intersection of the tangents to the curve $3x^2 - 6x + 2y^2 - 3 = 0$ from the point $T( 1,3 )$?
My work:
First (1,3) isn't on the curve. We will need that tangent equation: $$y-y_0=y'(x_0)(x-x_0)$$ From the task we know that $x=1$ and $y=3$.
Let's try to find $x_0$ and $y_0$. Using implict differentation we will find $y'$. $$6x-6+4yy'=0$$ $$3x-3+2yy'=0$$ $$2yy'=-3x+3$$ $$y'=\frac{-3x+3}{2y}$$ Plugin that in equation: $$3-y_0=\frac{-3x_0+3}{2y_0}(1-x_0)$$ $$3-y_0=\frac{-3x_0+3}{2y_0}+\frac{-3x_0^2-3x_0}{2y_0}$$ $$3-y_0=\frac{-3x_0^2+3-6x_0}{2y_0}$$ $$6y_0-2y_0^2=-3x_0^2+3-6x_0$$ But I can't get something nice...Is my problem 1 and 2 ok and please help me with 3

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  • $\begingroup$ The wording of problem (3) is odd: what does it really mean? "Intersection point of tanget to curve...from the point (1,3)..." Intersection of what with what? $\endgroup$ – DonAntonio Apr 3 '16 at 16:28
  • $\begingroup$ Tangents must cross point (1,3) and search angle of that tangents $\endgroup$ – josf Apr 3 '16 at 16:33
  • $\begingroup$ @Lo "angle..." with what? The positive $\;x-$ axis, as usual, or something else? $\endgroup$ – DonAntonio Apr 3 '16 at 16:38
  • $\begingroup$ @LovroSindičić: I edited the question. Hope that's fine. $\endgroup$ – mathlove Apr 3 '16 at 16:45
  • $\begingroup$ @LovroSindičić: In problem 3, you can draw two tangent lines from $(1,3)$ since the curve is an ellipse. $\endgroup$ – mathlove Apr 4 '16 at 11:28
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For problem 1,2 :

They look fine except the following :

One solution for $x\gt 0$ and $y\gt 0$ is $x=y=1$.

True, but that $(x,y)=(1,1)$ is the only solution for $x\gt 0,y\gt 0$ has to be proven. We have $y^2 =3x^2 −x−1$, and so $$x(3x^2 −x−1)−1=0⟺(x−1)\left(3\left(x+\frac 13\right)^2 +\frac 23\right)=0\iff x=1$$

For problem 3 :

You have $$3-y_0=\frac{-3x_0+3}{2y_0}(1-x_0),$$ i.e. $$-2y_0^2-3x_0^2+6x_0=-6y_0+3\tag1$$ Note here that $(x_0,y_0)$ is a point on the curve, so you have $$3x_0^2-6x_0+2y_0^2-3=0\iff -2y_0^2-3x_0^2+6x_0=-3\tag2$$ From $(1)(2)$, $$(x_0,y_0)=\left(\frac{3\pm 2\sqrt{3}}{3},1\right)$$

Hence, the slopes of the tangents are $$\frac{-3x_0+3}{2y_0}=\frac{-(3\pm 2\sqrt 3)+3}{2\cdot 1}=\mp\sqrt 3$$

Now find the angle in the similar way as problem 2.

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  • $\begingroup$ @Lovro Sindičić : For problem 3, after having $3-y_0=(-3x_0+3)(1-x_0)/(2y_0)$, you finally have $6y_0-2y_0^2=-3x_0^2+3-6x_0$. But this is wrong. You made a sign mistake at the second line. $\endgroup$ – mathlove Apr 3 '16 at 17:16
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Problem 1 is OK.

Problem 2 ;

First try to find intersection point $(x,y)$

$$ \frac1x - 3 x^2 + x +1 =0 $$

Factorize it

$$ -(x-1) (3 x^2 2 x +1) =0 $$

At $x=1, y= \dfrac{1}{\sqrt x } =1.\; i.e., (1,1) $ is the point of intersection. Other roots are neither in first quadrant nor real.

Now for the slope: $ y^2 = \frac 1x \rightarrow y'= \dfrac{-1}{ x^2 y} \rightarrow -1. $

Problem 3: not answered.

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The solutions to problems 1) and 2) are correct.

For problem 3) note that you are searching the straight lines of equation $(y-3)=m(x-1)$ that are tangent to the curve $3x^2-6x+2y^2-3=0$. This means that you want the values of $m$ such that the system: $$ \begin{cases} (y-3)=m(x-1)\\ 3x^2-6x+2y^2-3=0 \end{cases} $$ has one double solution, and this is done if the discriminant of the second degree equation that we obtain substituting from the first to the second equation is null. This gives an equation ( of second degree) in $m$ and solving this equation you find the slopes of the tangent lines.

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