Finding The Focus and Directrix of a Parabola - Conic Sections 

Full 1 Hour 42 Minute Video: www.patreon.com/MathScienceTutor Direct Link to The Full Video: bit.ly/3ju0COg

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Professor Organic Chemistry Tutor, thank you once again for another fantastic video/lecture on Finding the Focus, Directrix, Latus Rectum and Vertex of a Parabola. Professor Organic Chemistry Tutor, students all over the world should be watching your videos. The examples along with the graphs really defines a Parabola and its impact on Mathematics. This is an error free video/lecture on RU-vid TV with the Organic Chemistry Tutor.

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In the 4th question the eq of that graph is y²=-4px as it opens towards left so the answer is 12x

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14:35 the sign should be +. (y-k)^2 = 4p(x+h)

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24:13 the formula of focus should be (h,k-p) since its downward I memorized the formula already but this part got me confused when I tried to solve it, it is already determine that its downward but the formula you used is for upward it isn't facing to the positive it faces to the negative. Basically my answer for the focus is F=(3,2-(-2)) = +4, its downward not upward.

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Identify and label the vertex, focus, directrix, axis of symmetry, coordinates of latus rectum of the parabola given the equation (y-1)=-16x-48.

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26:13 Why the Domain is not (-1,7)? The FOCUS at point (3,0) so the Domain is (-1,7). Why(-infinity, +infinity)?

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I think the final answer in 13:05 was a positive y^2 = 12x. Cuz the parabola opens to the left and the p is negative then. The general that should be use is y^2= -4px. Right please correct me if I'm wrong though.

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But one thing ...... P cannot be negative cuz it's a line segment and is the length on the line so I guess the value can only be positive ......the negative sign indicates the direction/orientation or to where the graph opens.....could u plz elaborate on this part

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