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Q : Find the equation of the line that passes through the origin of the coordinations and the focus of the parabola $y=x^2+4x+1$.. so I found the focus $( -2;-3)$ and the line that passes through the origin is $y=x$..now what?

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3 Answers

$$\frac{y-0}{x-0}=\frac{0-(-3)}{0-(-2)}\implies 3x=2y$$

Also, any line passing through the origin is $y=mx$ where $m$ is the gradient which is $\frac{0-(-3)}{0-(-2)}=\frac 32$ here

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I dunno how you found $y=x$ but if the focus is indeed $(-2,-3)$ then the equation of the line joining the origin and the focus is $$\cfrac {y-0}{x-0}= \cfrac{-3-0}{-2-0}$$

EDIT:

Having checked, the focus you found is incorrect, the focus of a parabola $ax^2+bx+c=0$ is the point $$\left(-\cfrac{b}{2a}, \cfrac{1-b^2+4ac}{4a}\right)= \left(-2,-\cfrac{11}{4} \right)$$ And the equation of the line joining the focus$\left(-2,-\cfrac{11}{4} \right)$ and the origin $O(0,0)$ is $$\cfrac {y-0}{x-0}= \cfrac{(-11/4)-0}{-2-0} \implies y=\cfrac{11}{8}x$$

My guess is that you mistook the focus for the vertex $=\left(-\cfrac{b}{2a}, \cfrac{-b^2+4ac}{4a}\right)$

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You seem to have made an error, the focus is $(-2, -2.75)$. Therefore using the two point formula for a straight line passing through two points is

$ \frac{y - y_1}{x - x_1} = \frac{y_1 - y_2}{ x_1 - x_2} $

You can now substitute $(x_1,y_1)$ as $(0,0)$ and $(x_2,y_2)$ as $(-2,-2.75)$ to get the desired equation.

Ps: i'm not sure why you mention tangents here.

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