Coordinates on a parametric curve
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite 2 A curve is defined by the parametric equation $(x, y, z)$ $=$ $(-2 + 3t, 1 + 3t^2, 2t-3t^3)$ . There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(-8, 10, 8)$ . What are the coordinates of $P$ ? So I tried to find the equation of the tangent line first. So what I did was subtract the point $(-8, 10, 8)$ from the general equation of the line. This gave me $(-6-3t, 9-3t^2, 8-2t+3t^3)$ . The derivative for the general equation of the line is $(3 , 6t, 2-9t^2)$ . So I seem to understand that the tangent line is parallel, and therefore a scalar multiple, of the equation of the derivative. But I can't seem to see what the relationship is and how that can help me get the required coordinates. Any help? calculus derivatives parametric tangent-line share | cite | improve this question edited 2 hours ago Key Flex 6,023 1 8 28 ...