Ab is a chord of parabola y^2=4ax with its vertex at a.bc is drawn perpendicular to ab meeting the axis at c.the projection of bc on the axis of parabola is

Printable View

- Oct 13th 2010, 11:54 PMprasumparabola problem
Ab is a chord of parabola y^2=4ax with its vertex at a.bc is drawn perpendicular to ab meeting the axis at c.the projection of bc on the axis of parabola is

- Oct 14th 2010, 03:53 AMHallsofIvy
I don't understand what this means. "its" can only refer to the chord or the parabola. A chord does not have a vertex and the vertex of the parabola y^2= 4ax is at (0, 0).

Quote:

bc is drawn perpendicular to ab meeting the axis at c.the projection of bc on the axis of parabola is

- Oct 14th 2010, 05:11 AMprasum
the point 0,0 is A

- Oct 16th 2010, 08:32 AMHallsofIvy
So you mean "AB", not "Ab", "BC", not "bc", and the vertex is at "A" not "a". That's particularly confusing since you also used "a" as a number in the equation of the parabola.

If B is a point on the parabola, then $\displaystyle B= \left(\frac{y_0^2}{4a}, y_0\right)$ for some $\displaystyle y_0$ and the slope of AB is [tex]\frac{y}{\frac{y^2}{4a}}= \frac{4a}{y_0}[/quote]. Since BC is to be perpendicular to that, it must have slope $\displaystyle -\frac{y_0}{4a}$ and it passes through $\displaystyle B= \left(\frac{y_0^2}{4a}, y_0\right)$.

Write the equation of that line and set y= 0 in it to find x in C= (x, 0). The "projection" of of BC on the x-axis is the segment from $\displaystyle \left(\frac{y^2}{4a}, 0\right)$ to (x, 0).