Find tangent to a curve that pass through the origin (implicit function)

Question

I am trying to find the number of tangents to a curve that all pass through the origin.
The curve's equation is $y=x^3+x^2−22x+20.$ I also need to find the equation of said tangents.
My work:
Let's use formula for tangent line: $$y-y_0=y'(x_0)(x-x_0)$$ We know that $x=0$ and $y=0$.
Let's find $y'(x_0)$: $y'(x_0)=3x_0^2+2x_0-22$
Plugin what we know: $$-y_0=(3x_0^2+2x_0-22)*-x_0$$ We aslo know that $y_0=x_0^3+x_0^2-22x_0+20$ $$-x_0^3-x_0^2+22x_0-20=-3x_0^3-2x_0^2+22x_0$$ $$2x_0^3+x_0^2-20=0$$ Only real solution is 2. $x_0=2$ and $y_0=-12$. Let tangent line be $y=kx+l$.
$k=y'(2)=-6$
$-12=-6*2+l\to l=0$ But we want l to be zero. Our solution is $y=-6x$. Ok, that was an easy problem but what about equaly problem by my function look like this: $x^2-xy+y^2-1=0$. My question is how to solve equaly problem but my function is implicitly and I can't easily solve for $y$

Answer 1

what about equaly problem by my function look like this: $x^2-xy+y^2-1=0$. My question is how to solve equaly problem but my function is implicitly and I can't easily solve for $y$

You want to find the number of tangents to a curve $x^2-xy+y^2-1=0$ that all pass through the origin.

Let $(s,t)$ be the tangent point on the curve, i.e. $$s^2-st+t^2-1=0\tag1$$

We have $$2x-y-xy'+2yy'=0\iff (-x+2y)y'=-2x+y$$ Suppose that $-x+2y=0$, then $-2x+y=0$. Hence, $x=y=0$ but this point is not on the curve. Hence, $-x+2y\not=0$.

So a line tangent at $(s,t)$ is given by $$y-t=\frac{-2s+t}{-s+2t}(x-s)$$ Setting $x=y=0$ gives $$0-t=\frac{-2s+t}{-s+2t}(0-s),$$ i.e. $$s^2+t^2-st=0\tag2$$ However, there is no $(s,t)$ satisfying both $(1)$ and $(2)$. Hence, there is no such tangent.

---

In general, you would have two equations on two variables.

Answer 2

$$y=x^3+x^2-22x+20$$

$$y'=3x^2+2x-22$$

WLOG any point of the $y=x^3+x^2-22x+20$ can be $t,t^3+t^2-22t+20$

The equation of the tangent at $t,$ will be $$\dfrac{y-(t^3+t^2-22t+20)}{x-t}=3t^2+2t-22$$

Put $x=y=0$ and solve for $t$