Question

a circle of radius r drawn on a chord of the parabola y^2 =4ax as a diameter touches the axis of the parabola. prove that the slope of the chord is 2a/r

Answer 1

General point on a parabola with eq y2 =4ax  is  (at2 ,2at)

But we have eq y2 =4x

So a =1

So general point is (t2 ,2t) ..............1

Parabola cuts the circle at two different points ,let these points be A,B..

A=(t12,2t1)  & B=(t22,2t2)

Circle touches the x axis so the y coordinate of its center is +r when circle is drawn above x axis & -r when drawn below..

Since AB is diameter then

Coordinate of center of circle=A+B/2=[(t12 +t22)/2 ,(t1+t2)]

y coordinate of center =+ (r) or -(r)

So t1+t2=+ (r) or -(r)         .................2

Slope of AB=2(t2-t1)/(t22-t12)

 m=2/(t1+t2)

m=2/r ,-2/r

Answer 2

Answer:

General point on a parabola with eq y2 =4ax  is  (at2 ,2at)

But we have eq y2 =4x

So a =1

So general point is (t2 ,2t) ..............1

Parabola cuts the circle at two different points ,let these points be A,B..

A=(t12,2t1)  & B=(t22,2t2)

Circle touches the x axis so the y coordinate of its center is +r when circle is drawn above x axis & -r when drawn below..

Since AB is diameter then

Coordinate of center of circle=A+B/2=[(t12 +t22)/2 ,(t1+t2)]

y coordinate of center =+ (r) or -(r)

So t1+t2=+ (r) or -(r)         .................2

Slope of AB=2(t2-t1)/(t22-t12)

 m=2/(t1+t2)

m=2/r ,-2/r

Step-by-step explanation: