algebra-precalculuscalculuscircles
We're given that $x^2 + (y+2)^2 = 4$ and we're asked to find the equation of the lines where the gradient $=1$
Through implicit differentiation I got $x + (y+2)y'=0$
and if $y'=1$ then:
$y=-x-2$ is the relation where $y'=1$, which makes no sense to me.
I'm missing on something, so how do you go about finding the equations? (I know there're two.)
Since the two pairs of tangents are symmetric with respect to the $y$-axe, the quadratic function $f(x)=ax^{2}+bx+c$ must be even ($f(x)=f(-x)$), which implies that $b=0$. The equations of the tangents to the graph of $f(x)=ax^{2}+c$ at points $% \left( x_{1},f(x_{1})\right) $ and $\left( x_{2},f(x_{2})\right) $ are $$\begin{eqnarray*} y &=&f^{\prime }(x_{i})x-f^{\prime }(x_{i})x_{i}+f(x_{i})\qquad i=1,2 \\ &=&2ax_{i}x+c-ax_{i}^{2}. \end{eqnarray*}$$
These equations must be equivalent to two of the given tangents, one from each pair, e.g. $y=2x-10$ and $y=x-4$:
$$\left\{ \begin{array}{c} 2ax_{1}x+c-ax_{1}^{2}=2x-10 \\ 2ax_{2}x+c-ax_{2}^{2}=x-4% \end{array}% \right. $$
Finally we compare coefficients and solve the resulting system of $4$ equations:
$$\left\{ \begin{array}{c} 2ax_{1}=2 \\ c-ax_{1}^{2}=-10 \\ 2ax_{2}=1 \\ c-ax_{2}^{2}=-4% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} x_{1}=8 \\ x_{2}=4 \\ a=\frac{1}{8} \\ c=-2% \end{array}% \right. $$
Thus the quadratic is $f(x)=\frac{1}{8}x^{2}-2$.
Solving for gradient [i.e. "slope"] at $x = 0$ ($x = 0$ is that is the equation of the y-axis) given
$\;y^3+2y = \sin x+\cos x-1+3x^{2}\tag{1}$
Solve for $y$ at $x = 0$: $\quad x = 0 \implies y^3 + 2y = y(y^2 + 2) = 0\implies y = 0, \;x, y \in \mathbb R$
Use implicit differentiation on the left hand side and the right hand side: as a check, $$\frac{dy}{dx} = \frac{6x + \cos x - \sin x}{3y^2 + 2}$$
Evaluating the derivative at $(0, 0)$ will give the slope/gradient $m$ of the curve at the $y\text{-intercept}.\;$ As a check: we should get $m = 1/2.$
And here, in fact, is the graph of your function near the origin, and a graph of your function with the line tangent to the curve at $(0, 0)$
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Best Answer
the equation $x^2 + (y+2)^2 = 4$ is the graph of the circle centered at $(0, -2)$ and of radius $2.$ tangents at the end of the diameter joining the points $(\sqrt 2, -2 - \sqrt 2), (-\sqrt 2, -2 + \sqrt 2)$ has the slope $1.$ they are given by $$x - y = \pm(2\sqrt 2 - 2).$$