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

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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:

$y=x-1,\;y=x+\dfrac{13}{3}$

#### Prove It

##### Well-known member
MHB Math Helper
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*}.