1. parabola 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

2. Originally Posted by prasum
Ab is a chord of parabola y^2=4ax with its vertex at a.
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).

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

3. the point 0,0 is A

4. 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 $B= \left(\frac{y_0^2}{4a}, y_0\right)$ for some $y_0$ 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 $-\frac{y_0}{4a}$ and it passes through $B= \left(\frac{y_0^2}{4a}, y_0\right)$.

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 $\left(\frac{y^2}{4a}, 0\right)$ to (x, 0).