## Mimi_x3 3 years ago 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

1. Ishaan94

$x^2 = 4ay$ |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

2. Mimi_x3

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

3. Ishaan94

'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

4. Ishaan94

ah Okay I think I got it

5. Ishaan94

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

6. Kira_Yamato

|dw:1319020480071:dw|

7. Kira_Yamato

|dw:1319020633975:dw|

8. Kira_Yamato

|dw:1319020681433:dw|

9. Ishaan94

equation of normal must be according to the y = mx + c form $y' = \frac{-2a}{x}x' + c$Now we have points (x,y) for the Normal on parabola $$x^2 = 4ay$$ $y = \frac{-2a}{x}x+c$ $\frac{x^2}{4a} = -2a + c$ $\frac{x^2}{-8a^2} = c$

10. Ishaan94

sorry, typo ignore the last expression

11. Ishaan94

$\frac{x^2 + 2a*4a}{4a}=c$ $\frac{x^2 + 8a^2}{4a}=c$

12. Ishaan94

OM is from origin perpendicular to PK equation of OM $y' = \frac{x}{2a}x'$

13. Ishaan94

equation of PK $y' = x' \frac{-2a}{x} + \frac{x^2 + 8a^2}{4a}$

14. Ishaan94

The point M satisfies both the equation i.e equation of PK and equation of OM

15. Ishaan94

Let M be (x_1,y_1) $y_1 = x_1* \frac{-2a}{x} + \frac{x^2 + 8a^2}{4a}$ $y_! = x_1 *\frac{x}{2a}$ ${x_1*}\frac{-2a}{x} + \frac{x^2 + 8a^2}{4a} = \frac{x}{2a}*x_1$ $x_1 \left( \frac{4a^ 16. Ishaan94 \[x_1 \left( \frac{4a^2 + x^2 }{2ax}\right) = \frac{x^2 + 8a^2}{4a}$

17. Ishaan94

$x_1 = \frac{x^3 + 8xa^2}{8a^2 + 2x^2}$

18. Ishaan94

$y_1 = \frac{x}{2a}\times \frac{x^3 + 8a^2x}{8a^2 +2x^2}$

19. Mimi_x3

I'm lost =/ Isn't the equation of the normal y = -x + 2ap^3/p

20. Ishaan94

I don't think so $x^2 = 4ay$ $\frac{dy}{dx} = \frac{x}{2a}$ Slope of tangent, Normal is perpendicular to tangent

21. Ishaan94

This is getting complicated, I will try this later I think I am getting the question wrong. I will try this on my notebook first

22. Mimi_x3

It's okay never mind then, I will ask my teacher tomorrow.