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Mimi_x3
 5 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
Mimi_x3
 5 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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2 = 4ay\] 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
 5 years ago
Best ResponseYou've already chosen the best response.3lols, im confused as well, i've attempted on drawing it but i think that it's wrong =/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0'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
 5 years ago
Best ResponseYou've already chosen the best response.0ah Okay I think I got it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe the figure is like thisdw:1319019992221:dw

Kira_Yamato
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1319020480071:dw

Kira_Yamato
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1319020633975:dw

Kira_Yamato
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1319020681433:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0equation 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 \(x^2 = 4ay\) \[y = \frac{2a}{x}x+c\] \[\frac{x^2}{4a} = 2a + c \] \[ \frac{x^2}{8a^2} = c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, typo ignore the last expression

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x^2 + 2a*4a}{4a}=c\] \[\frac{x^2 + 8a^2}{4a}=c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OM is from origin perpendicular to PK equation of OM \[y' = \frac{x}{2a}x'\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0equation of PK \[y' = x' \frac{2a}{x} + \frac{x^2 + 8a^2}{4a} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The point M satisfies both the equation i.e equation of PK and equation of OM

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let 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^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x_1 \left( \frac{4a^2 + x^2 }{2ax}\right) = \frac{x^2 + 8a^2}{4a}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x_1 = \frac{x^3 + 8xa^2}{8a^2 + 2x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y_1 = \frac{x}{2a}\times \frac{x^3 + 8a^2x}{8a^2 +2x^2} \]

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.3I'm lost =/ Isn't the equation of the normal y = x + 2ap^3/p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think so \[ x^2 = 4ay \] \[\frac{dy}{dx} = \frac{x}{2a}\] Slope of tangent, Normal is perpendicular to tangent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is getting complicated, I will try this later I think I am getting the question wrong. I will try this on my notebook first

Mimi_x3
 5 years ago
Best ResponseYou've already chosen the best response.3It's okay never mind then, I will ask my teacher tomorrow.
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