prove that circle drawn with focal chord of parabola as diameter touches directrix..........?

sorry silly mistake

Let d parabola b Y² = 4ax

d end pts of focal chord wud b (a,2a) & (a,-2a)



D eq of circle with d above pts as diameter wud b :

(x-a)(y-2a) + (x-a)(y+2a) = 0



Now d eq of directrix wud b x = -a

To find out d pts of intersection of directrix & circle,

subs x = -a in d eq of circle. U wud get d ans as y = 0



so this means tat d line x= -a touches d circle at xactly 1 pt = (-a,0)

Hence proved tat d directrix of d parabola is a tangent 2 d circle.



Dont forget 2 rate & feel free 2 ask ne more of ur doubts.

U can try d above by taking d parabola as x² = 4ay - wudnt make ne diff.

Consider tat here 'a' can b +ve or -ve, so tat takes up all d cases

Plz note maxzy focal chord and latus rectum are not same.A chord to the parabola passing thro' focus is called a focal chord.

Now,let us assume the vertex of the parabola as origin and its axis as x-axis.Then it's eqn. becomes y²=4ax with focus (a,0) and directrix x+a=0.

Let P(t),Q(t') be the extremities of the focal chord then tt'=-1.So,P=(at²,2at) and Q=(a/t²,-2a/t).

PQ²=a²(t+1/t)⁴PQ=a(t+1/t)² and so,radius R of the circle=(a/2)(t+1/t)².

Now,the circle drawn with PQ as diameter touches the directrix iff

Distance d of the directrix is equal to radius R i.e;d=R.

Here,Centre of circle(x₀,y₀)=((a/2)(t²+1/t²),a(t-1/t))

and so,d=Ix₀+aI=(a/2)(t²+1/t²)+a=(a/2)[t²+1/t²+2]=(a/2)(t+1/t)²=R.

Hence,the directrix touches the given circle.