MHB Bruno's question at Yahoo Answers (Tangent line to an ellipse).

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Ellipse Line
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Here is the question:

3x^2+y^2+4x-2y=3 Find the equation of the tangent line to that ellipse with slope 1.
I know how to write it in the canonic form but then i don't know what to do.
Answer: x-y-1=0

Here is a link to the question:

Tangent line to a ellipse? - Yahoo! Answers
 
Mathematics news on Phys.org
Hello Bruno,

The equation of a line with slope $1$ is $r:y=x+b$. If the line is tangent to the ellipse, the intersection must be only one point. Substituting in the ellipse:

$3x^2+(x+b)^2+4x-2(x+b)-3=0$

Equivalently:

$4x^2+(2b+2)x+b^2-2b-3=0$

The discriminat $D=B^2-4AC$ must be $0$:

$D=(2b+2)^2-16(b^2-2b-3)=0$

Simplifying:

$3b^2-10b-13=0\Leftrightarrow b=-1\vee b=13/3$

So, there are two solutions:

$y=x-1,\;y=x+\dfrac{13}{3}$
 
Fernando Revilla said:
Here is the question:
Here is a link to the question:

Tangent line to a ellipse? - Yahoo! Answers

An alternative method:

Your tangent line is $\displaystyle \begin{align*} 3x^2 + y^2 + 4x - 2y = 3 \end{align*}$. Differentiating both sides with respect to x gives

$\displaystyle \begin{align*} \frac{d}{dx} \left( 3x^2 + y^2 + 4x - 2y \right) &= \frac{d}{dx} \left( 3 \right) \\ 6x + 2y\,\frac{dy}{dx} + 4 - 2\,\frac{dy}{dx} &= 0 \\ 6x + 4 &= \left( 2 - 2y \right) \frac{dy}{dx} \\ 3x + 2 &= \left( 1 - y \right) \frac{dy}{dx} \\ \frac{3x + 2}{1 - y} &= \frac{dy}{dx} \end{align*}$

You know that the slope is 1 at that point, so that means

$\displaystyle \begin{align*} \frac{3x + 2}{1 - y} &= 1 \\ 3x + 2 &= 1 - y \\ 3x + 1 &= -y \\ -3x - 1 &= y \end{align*}$

Substituting into the original equation for your ellipse gives

$\displaystyle \begin{align*} 3x^2 + y^2 + 4x - 2y &= 3 \\ 3x^2 + \left( -3x - 1 \right)^2 + 4x - 2 \left( -3x - 1 \right) &= 3 \\ 3x^2 + 9x^2 + 6x + 1 + 4x + 6x + 2 &= 3 \\ 12x^2 + 16x + 3 &= 3 \\ 12x^2 + 16x &= 0 \\ 4x \left( 3x + 4 \right) &= 0 \\ 4x = 0 \textrm{ or } 3x + 4 &= 0 \\ x = 0 \textrm{ or } x &= -\frac{4}{3} \end{align*}$

And since we know $\displaystyle \begin{align*} y = -3x - 1 \end{align*}$, that means the two points on the ellipse which have a tangent line of slope 1 are $\displaystyle \begin{align*} (0, -1) \end{align*}$ and $\displaystyle \begin{align*} \left( -\frac{4}{3} , 3 \right) \end{align*}$.

Each tangent line will be of the form $\displaystyle \begin{align*} y = mx + c \end{align*}$, so in the first tangent line:

$\displaystyle \begin{align*} -1 &= 1(0) + c \\ -1 &= c \end{align*}$

and in the second tangent line:

$\displaystyle \begin{align*} 3 &= 1 \left( -\frac{4}{3} \right) + c \\ 3 &= -\frac{4}{3} + c \\ \frac{13}{3} &= c \end{align*}$So your two tangent lines are $\displaystyle \begin{align*} y = x - 1 \end{align*}$ and $\displaystyle \begin{align*} y = x + \frac{13}{3} \end{align*}$.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top