parametrics 2 (1 Viewer)

jnney · Oct 1, 2011

the normal at any point P(-8p, -4p^2) on the parabola x^2=-16y cuts the y-axis at thepoint M. find the equation of the locus of the midpoint of PM.

unknown88 · Oct 1, 2011

jnney said:
the normal at any point P(-8p, -4p^2) on the parabola x^2=-16y cuts the y-axis at thepoint M. find the equation of the locus of the midpoint of PM.

y = x^2 / -16 ... dy/dx = -x/8 ...sub x= -8p. so dy/dx = p = GRADIENT of P ...

y - y-cord. of P = p(gradient) [ x- x-cord of p]

y + 4p^2 = p ( x+8p)

y + 4p^2 = px +8p^2
y = px + 4p^2... Thats the equation...the question says M cuts the y-axis, therefore x=0 so sub that into equation

y = p(0) + 4p^2
y=4p^2 so M (0,4p^2) .. just find the midpoint formula and sub in values.

unknown88 · Oct 1, 2011

reply wether my method was right or wrong

jnney · Oct 1, 2011

unknown88 said:
y = x^2 / -16 ... dy/dx = -x/8 ...sub x= -8p. so dy/dx = p = GRADIENT of P ...

y - y-cord. of P = p(gradient) [ x- x-cord of p]

y + 4p^2 = p ( x+8p)

y + 4p^2 = px +8p^2
y = px + 4p^2... Thats the equation...the question says M cuts the y-axis, therefore x=0 so sub that into equation

y = p(0) + 4p^2
y=4p^2 so M (0,4p^2) .. just find the midpoint formula and sub in values.

Mhm, the normal, not the tangent.

unknown88 · Oct 1, 2011

then if its normal change the gradient to -1 / p and go just go thru my process agen

jnney · Oct 1, 2011

unknown88 said:
then if its normal change the gradient to -1 / p and go just go thru my process agen

Yes I did that before posting up the question. I became confused once I had found the equation.

nightweaver066 · Oct 1, 2011

$x^2 = -16y$

$y = -\frac{x^2}{16}$

$\frac{dy}{dx} = -\frac{x}{8}$

$\text{At }P(-8p, -4p^2), \frac{dy}{dx} = p$

$\therefore m(normal) = -\frac{1}{p}$

$\text{Equation of normal:}$

$y + 4p^2 = -\frac{1}{p}(x + 8p)$

$\text{At x = 0, } y + 4p^2 = -8$

$y = -4p^2 - 8 => M(0, -4p^2 - 8)$

$\text{Let R(x,y) be a point in the plane}$

$\text{For } RP = RM$

$\sqrt{(x+8p)^2 + (y + 4p^2)^2} = \sqrt{x^2 + (y + 4p^2 + 8)^2}$

$x^2 + 16px + 64p^2 + y^2 + 8p^2y + 16p^4 = x^2 + y^2 + 4p^2y + 8y + 4p^2y + 16p^4 + 32p^2 + 8y + 32p^2 + 64$

$16px + 64p^2 + 8p^2y + 16p^4 =8p^2y + 16y + 16p^4 + 64p^2 + 64$

$16px + 16y - 64 = 0$

$px + y - 4 = 0$

bleakarcher · Oct 2, 2011

x^2+4y+16=0

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