Points of intersections of the line y=mx+c with the parabola y2=4ax:
Three Possibilities:
To find the points of intersection y=mx+c and y2=4ax, we solve
(mx+c)2=4ax
⇔m2x2+2mcx+c2=4ax
⇔m2x2+2(mc−2a)x+c2=0
This is a quadratic equation in x, so it has either two real & distinct roots or equal roots or two non real complex roots.
Disriminant = [2(mc−2a)]2−4m2c2
= 4[m2c2−4amc+4a2−m2c2]
= 16(a2−amc)
∴ there are 2 points of intersection if a2−amc>0
that is a>mc
There is one point of intersection if a=mc
& No points if intersection if $a
When a=mc the line y=mx+c intersects the parabola is only one point, and hence it is tangent to the parabola y2=4ax at the point (x,y) given by
x=2m2−2(mc−2a)=2m22(mc−2mc)
=m2mc=mc
Putting x=mc in the equation. y=mx+c gives y=m(mc)+2c
So, the line y=mx+c is tangent to the parabola y2=4ax at the point (mc,2c) provided mc=a
Length of chord of parabola y2=4ax:
Length of the chord joining (x1,y2) & (x2,y2) on the parabola y2=4ax is
l = (x1−x2)2+(y1−y2)2
Equation of the line joining (x1,x2) & (x2,y2) is
y−y1=x1−x2y1−y2(x−x1)
y=mx+c, where m=x1−x2y1−y2
c=y1−(x1−x2y1−y2)x1
Provided x1=x2
Note that if x1=x2 then
l=∣y1−y2∣
y12=4ax,y22=4ax
y1=−y2 & ∣y1−y2∣=24ax1=4ax1
x1 and x2 satisfy the quadratic equation.
m2x2+2(mc−2a)x+c2=0
∴x1+x2=m2−2(mc−2a)
& x1x2=m2c2
∴(x1−x2)2=(x1+x2)2−4x1x2
=m44(mc−2a)2−m24c2
=m44[m2c2−4amc+4a2−m2c2]
=m416a(a−mc)
Also, y1−y2=(mx1+c)−(mx2+c)=m(x1−x2)
∴ Length of the chord, l = (x1−x2)2+(y1−y2)2
= (x1−x2)2+m2(x1−x2)2
= 1+m2∣x1−x2∣
= 1+m2m24a(a−mc)
To get the formula for l in terms of (x1,x2) & (y1,y2) we can put
m=x1−x2y1−y2 & c=y1−(x1−x2y1−y2)x1
Equation of the tangent line to the parabola y2=4ax at a point (x1,x2):
Suppose the line has slope m then the equation of the tangent line is given by y−y1=m(x−x1)
i.e y=mx+(y1−mx1)
We know that this line is tangent to y2=4ax
if a=mc=m(y1−mx1)
⇔m2x1−my1+a=0
⇔m=2x1y1±y12−4ax1, But y12=4ax1
So, m=2x1y1 This can also be written as
m=ax1y1y12=ax1y14ax1=y12a
m=y12a
Equation of the tangent line is
y=mx+(y1−mx1)
= y12ax+y1−2y1
= y12ax+2y1
⇔yy1=2ax+2y2=2ax+24ax1
⇔yy1=2a(x+x1)
Equation of the tangent line to the parabola y2=4ax at the point (x1,y1) on the parabola.
y2=4ax: Equation of the parabola (x1,y1) lies on the parabola ⇒y12=4ax1. we know that the slope of the tangent line at a point (x1,y1) on any curve y=f(x) is given by m=dxdy∣e(x1,y1)
y2=4ax⇒2ydxdy=4a
⇒dxdy=y2a⇒m=y12a
∴ equation of the tangent line is
y−y1=y12a(x−x1)
⇔yy1−y12=2ax−2ax1
But y12=4ax1
∴yy1−4ax1=2ax−2ax1
⇔yy1=2ax+2ax1
⇔yy1=2a(x+x1)
Remark: The above derivation assumed
(x1,y1)=0
However, if the point (x1,y1)=(0,0) Then it is clear that the tangent line at (0,0) is the y-axis, where equation is x=0.
If we put (x1,y1)=(0,0) in the equation obtained yy1=2a(x+x1)
we get 0 = 2a(x+0) i.e x=0
Hence, the equation yy1=2a(x+x1) is valid for even the point (0,0)
general equation of tangent at (x1,y1)
Equation of the normal at point (x1,y1) on the parabola y2=4ax:
Equation of the tangent is yy1=2a(x+x1)
If y1=0 then y=y12a(x+x1)
∴ Slope of tangent = y12a
⇒ Slope of normal, m = 2a−y1
∴ Equation of normal is
y−y1=2a−y1(x−x1)
In terms of the slope m:
m=2a−y1 i.e y1=−2am
∴x1=4ay12=4a4a2m2=am2
Putting in the equation. y−y1=2a−y1(x−x1),
we get
y+2am=m(x−am2)
⇔y=mx−2am−am3
equation of the normal at the points (x1,y1)=(am2,−2am)
AT: subtangent
AN: subnormal
Equation of line PT is
yy1=2a(x+x1)
Putting y=0 gives x=−x1
∴T≡(−x1,0)
Subtangent AT=2x1
△TAP is similar to △PAN
⇒APAN=ATAP⇒AN=ATAP2=2x1y12=2x14ax1=2a
∴ Subnormal, AN=2a
Which is a constant