$P$ $P$ is a point on the parabola $x = t$ $x=t$ , $y = \frac{t^{2}}{2}$ $y=t^2/2$ . $A (4, 1)$ $A(4,1)$ is a fixed point. As $P$ $P$ varies, find the minimum distance of $P$ $P$ from $A$ $A$ and prove that for this position of $P$ $P$ , $A P$ $AP$ is normal to the parabola?

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$P$ $P$ is a point on the parabola $x = t$ $x=t$ , $y = \frac{t^{2}}{2}$ $y=t^2/2$ . $A (4, 1)$ $A(4,1)$ is a fixed point. As $P$ $P$ varies, find the minimum distance of $P$ $P$ from $A$ $A$ and prove that for this position of $P$ $P$ , $A P$ $AP$ is normal to the parabola?

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Answer 1 · 2017-11-20T14:12:00

The variable distance $A P = s$ $AP=s$ is given by

$s = \sqrt{{(t - 4)}^{2} + {(\frac{t^{2}}{2} - 1)}^{2}}$ $s=sqrt((t-4)^2+(t^2/2-1)^2$

$\Rightarrow s^{2} = t^{2} - 8 t + 16 + \frac{t^{4}}{4} - t^{2} + 1$ $=>s^2=t^2-8t+16+t^4/4-t^2+1$

$\Rightarrow s^{2} = \frac{t^{4}}{4} - 8 t + 17$ $=>s^2=t^4/4-8t+17$

Differentiating w r to t we get

$2 s \frac{d s}{d t} = t^{3} - 8$ $2s(ds)/(dt)=t^3-8$

Imposing the condition of minimum $\frac{d s}{d t} = 0$ $(ds)/(dt)=0$ we get $t^{3} = 8 \Rightarrow t = 2$ $t^3=8=>t=2$

At this position the coordinates of $P$ $P$ becomes $(2, \frac{2^{2}}{2}) or (2, 2)$ $(2,2^2/2) or (2,2)$

At this stage the slope of $A P$ $AP$ will be $\frac{2 - 1}{2 - 4} = - \frac{1}{2}$ $(2-1)/(2-4)=-1/2$

Now the equation of the parabola $x^{2} = 2 y$ $x^2=2y$

So $\frac{d y}{d x} = x$ $(dy)/(dx)=x$

Hence slope of the normal at $(2, 2)$ $(2,2)$ will be

$= - \frac{1}{{[\frac{d y}{d x}]}_{2, 2}} = - \frac{1}{2}$ $=-1/[(dy)/(dx)]_(2,2)=-1/2$ which is slope of $A P$ $AP$ .Hence $A P$ $AP$ is the normal to the parabola.

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$P$ $P$ is a point on the parabola $x = t$ $x=t$ , $y = \frac{t^{2}}{2}$ $y=t^2/2$ . $A (4, 1)$ $A(4,1)$ is a fixed point. As $P$ $P$ varies, find the minimum distance of $P$ $P$ from $A$ $A$ and prove that for this position of $P$ $P$ , $A P$ $AP$ is normal to the parabola?

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PPP is a point on the parabola x=tx=tx=t, y=t^2/2y=t22y=t^2/2. A(4,1)A(4,1)A(4,1) is a fixed point. As PPP varies, find the minimum distance of PPP from AAA and prove that for this position of PPP, APAPAP is normal to the parabola?

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$P$ $P$ is a point on the parabola $x = t$ $x=t$ , $y = \frac{t^{2}}{2}$ $y=t^2/2$ . $A (4, 1)$ $A(4,1)$ is a fixed point. As $P$ $P$ varies, find the minimum distance of $P$ $P$ from $A$ $A$ and prove that for this position of $P$ $P$ , $A P$ $AP$ is normal to the parabola?