parametrics (1 Viewer)

jnney · Oct 1, 2011

Find the equation of the locus of point R that is the intersection of the normals at P(2p, p^2) and Q(2q, q^2) on the parabola x^2=4y given that pq=-4

maths lover · Oct 1, 2011

do you understand the process involved or are you having troubles removing the parameters?.

SpiralFlex · Oct 1, 2011

If you are familiar with the process. I will skip it. I better derive it just incase, editing it now.

$y=\frac{x^2}{4}$

$\frac{dy}{dx}=\frac{x}{2}$

At $(2p, p^2)$

The gradient of the normal is $-\frac{1}{p}$

$py-p^2=-\frac{1}{p}(x-2p)$

$py-p^3 = -x+2p$

$x+py=p^3+2p$

Similarly,

$x+qy=q^3+2q$

The intersection of the normals are $(-pq(p+q), (p^2+q^2+pq+2))$

Let's write this for something more comfortable.

$x=-pq(p+q)$ (1)

$y=(p^2+q^2+pq+2)$ (2)

From 1,

$\frac{x}{4}=p+q$

From 2,

$y=(p+q)^2-pq+2$

$y=\frac{x^2}{16}+6$

$x^2=16(y-6)$

parametrics (1 Viewer)

jnney

lemon

maths lover

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SpiralFlex

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