At a point P on the parabola x^2=4ay, a normal PK is drawn. From the vertex O a perpendicular OM is drawn to meet the normal at M. Show that the equation of the locus of M as P varies on the parabola is x^4 -2ax^2y + x^2y^2 - ay^3 = 0

Question

anonymous · Answer

$$x^2 = 4ay$$
|dw:1319018817978:dw|
Differentiating with respect to x
$$2x = 4a \frac{dy}{dx}$$
$$\frac{x}{2a} =\frac{dy}{dx}$$
Normal is perpendicular  to tangent 
Hence, Slope of Normal is $-\frac{2a}{x}$

@mimi can you draw the figure? I am confused

mimi_x3 · Answer

lols, im confused as well, i've attempted on drawing it but i think that it's wrong =/

anonymous · Answer

'From the vertex O a perpendicular OM is drawn to meet the normal at M'


Maybe something is not right here, I am not sure but try reading the question again maybe some typo

anonymous · Answer

ah Okay I think I got it

anonymous · Answer

maybe the figure is like this|dw:1319019992221:dw|