Find the points on the ellipse where the slope of the tangent line is 1.

Question

Find the points on the Ellipse $x^2+2y^2=1$ where the tangent line has a slope 1. Need a refresh on a problem like this.

Answer 1

The slope of the tangent is $y'$

$$x^2+2y^2=1\implies2x+4yy' = 0 \implies y' = -\frac{x}{2y}=1 (y\ne0)\implies x=-2y, $$

Use this in $x^2+2y^2=1$

Answer 2

Differentiating with respect to $x$ we get $$2x+4yy'=0$$ so $$y'=-\frac{1}{2}\frac{x}{y}=1$$ Can you proceed? For $y\neq 0$

Answer 3

By implicit differentiation, $2x+4y\dfrac{dy}{dx}=0,$ so $\dfrac{dy}{dx}=1\implies x=-2y$.

Therefore, $x^2+2y^2=1\implies6y^2=1$. From there, it is easy to find $y$ and then $x$.

Answer 4

WLOG any point on the ellipse, $x=\cos t,\sqrt2y=\sin t$

$$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=-\dfrac{\cos t}{\sqrt2\sin t}$$ which needs to be $1$

$$\implies$$

$$\dfrac{\cos t}{-\sqrt2}=\dfrac{\sin t}1=\pm\sqrt{\dfrac{\cos^2t+\sin^2t}{?}}$$

Answer 5

In the ellipse: $a=1$ and $b=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$. The equation of the tangent line to the ellipse (with centre in $(0,0)$) is: $\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1$ where $(x_0,y_0)$ are the coordinates of the intersection. I can substitute and I obtain: $$\frac{xx_0}{1}+\frac{yy_0}{\frac{1}{2}}=1$$ So: $2yy_0=1-xx_0$ and $y=\frac{1}{2y_0}-\frac{xx_0}{2y_0}$. Substituting, I obtain: $$m=\frac{\sqrt{2}x_0}{2\sqrt{1-x_0^2}}$$ Solving $m=1$ I obtain: $(x_0=\pm\sqrt{\frac{2}{3}},y_0=\mp\sqrt{\frac{1}{6}})$. I hope it has helped you...

Answer 6

An equation of the tangent it's $$xx_1+2yy_1=1,$$ where $(x_1,y_1)$ is a touching point.

Thus, since the slope is equal to $1$ we obtain: $$x_1=-2y_1,$$ which gives $$4y_1^2+2y_1^2=1$$ and we got the answer: $$\left\{\left(-\sqrt{\frac{2}{3}},\frac{1}{\sqrt6}\right),\left(\sqrt{\frac{2}{3}},-\frac{1}{\sqrt6}\right)\right\}$$