Question #00f1a

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Answer 1 · 2017-04-07T16:41:16

The given equation of the parabola

$y=4ax^2$

$=>x^2=4xx1/(16a)y$

Hence the parabola is symmetric about y-axis i,e its axis is Y-axis .

The co-ordinates of its vertex is $=>(0,0)$ .

The co-ordinates of its focus ( F) $=>(0,1/(16a))$

The equation of its directrix, $=>y=-1/(16a)$

Let the co-ordinates of any point P on it in parametric form be $((2t)/(16a),t^2/(16a))$ ,where $t$ is the parameter.

Now slope of the tangent at P

$((dy)/(dx))_(((2t)/(16a),t^2/(16a))) =4axx2xx(2t)/(16a)=t$

So slope of the normal at P $=-1/t$

So equation of the normal at P

$y-t^2/(16a)=-1/txx(x-(2t)/(16a))$

Putting , $y=-1/(16a)$ in the equation of the normal we get the X=coordinate of the point of intersection (Q) of the normal with the directrix

$-1/(16a)-t^2/(16a)=-1/txx(x-(2t)/(16a))$

$=>-1/(16a)-t^2/(16a)=-x/t+2/(16a)$

$=>x/t=t^2/(16a)+3/(16a)$

$=>x=t^3/(16a)+(3t)/(16a)$

So co-ordinates of Q is $(t^3/(16a)+(3t)/(16a),-1/(16a))$

So

$PQ^2=[(t^3/(16a)+(3t)/(16a)-(2t)/(16a))^2+(t^2/(16a)+1/(16a))^2]$

$=1/(16a)^2[(t^3+t)^2+(t^2+1 )^2]$

$=1/(16a)^2[t^2(t^2+t)^2+(t^2+1 )^2]$

$=1/(16a)^2[(t^2+t)^2(t^2+1 )]$

$=1/(16a)^2[(t^2+t)^3]$

No

$FP^2=((2t)/(16a))^2+(t^2/(16a)-1/(16a))^2=(t^2+1)^2/(16a)^2$

$=>FP^3=(t^2+1)^3/(16a)^3$

Hence

$(PQ^2)/(FP^3)=16a$

$=>PQ^2=16aFP^3$