The normal $(2 a p, a p^{2})$ $(2ap, ap^2)$ to the parabola $x^{2} = 4 a y$ $x^2=4ay$ meets the curve again at $Q (2 a q, a q^{2})$ $Q(2aq, aq^2)$ ?

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The normal $(2 a p, a p^{2})$ $(2ap, ap^2)$ to the parabola $x^{2} = 4 a y$ $x^2=4ay$ meets the curve again at $Q (2 a q, a q^{2})$ $Q(2aq, aq^2)$ ?

(a) Show that $q = - \frac{2 + p^{2}}{p}$ $q=-(2+p^2)/p$

(b) Find the coordinates of P so that the lines $O P$ $OP$ and $O Q$ $OQ$ meet at right angles, where $O$ $O$ is the origin.

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Answer 1 · 2017-11-20T13:17:56

(a)Given equation of the parabola
$x^{2} = 4 a y$ $x^2=4ay$

$\Rightarrow \frac{d y}{d x} = \frac{1}{4 a} \cdot 2 x = \frac{x}{2 a}$ $=>(dy)/(dx)=1/(4a)*2x=x/(2a)$

Slope of the normal

$m = - \frac{1}{{[\frac{d y}{d x}]}_{2 a p, a p^{2}}^{2}} = - \frac{2 a}{2 a p} = - \frac{1}{p}$ $m=-1/[(dy)/(dx)]_(2ap,ap^2)=-(2a)/(2ap)=-1/p$

Again the slope of the normal as it joins $(2 a p, a p^{2}) and (2 a q, a q^{2})$ $(2ap,ap^2) and (2aq,aq^2)$ is also
$m = \frac{a p^{2} - a q^{2}}{2 a p - 2 a q} = \frac{1}{2} (p + q)$ $m=(ap^2-aq^2)/(2ap-2aq)=1/2(p+q)$

So we have

$\frac{1}{2} (p + q) = - \frac{1}{p}$ $1/2(p+q)=-1/p$

$=>q=-2/p-p=-(2+p^2)/p....(1)$

OP and OQ being orthogonal the produt of their slopes $(ap^2)/(2ap)*(aq^2)/(2aq)=-1$

$=>pq=-4....(2)$

From (1) and (2) we get

$-4=-(2+p^2)$

$=>p=sqrt2$

(b)So coordinates of the point P will be $(2sqrt2a,2a)$

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The normal $(2 a p, a p^{2})$ $(2ap, ap^2)$ to the parabola $x^{2} = 4 a y$ $x^2=4ay$ meets the curve again at $Q (2 a q, a q^{2})$ $Q(2aq, aq^2)$ ?

(a) Show that $q = - \frac{2 + p^{2}}{p}$ $q=-(2+p^2)/p$

(b) Find the coordinates of P so that the lines $O P$ $OP$ and $O Q$ $OQ$ meet at right angles, where $O$ $O$ is the origin.

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The normal (2ap, ap^2)(2ap,ap2)(2ap, ap^2) to the parabola x^2=4ayx2=4ayx^2=4ay meets the curve again at Q(2aq, aq^2)Q(2aq,aq2)Q(2aq, aq^2)?

(a) Show that q=-(2+p^2)/pq=−2+p2pq=-(2+p^2)/p (b) Find the coordinates of P so that the lines OPOPOP and OQOQOQ meet at right angles, where OOO is the origin.

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The normal $(2 a p, a p^{2})$ $(2ap, ap^2)$ to the parabola $x^{2} = 4 a y$ $x^2=4ay$ meets the curve again at $Q (2 a q, a q^{2})$ $Q(2aq, aq^2)$ ?

(a) Show that $q = - \frac{2 + p^{2}}{p}$ $q=-(2+p^2)/p$

(b) Find the coordinates of P so that the lines $O P$ $OP$ and $O Q$ $OQ$ meet at right angles, where $O$ $O$ is the origin.