

Parabola
and Line

Common points of a line and a parabola 
Condition for a line to be the tangent to the parabola
 tangency condition

The equation of the tangent and the normal at the point on the parabola

Parabola and line, examples 





Common points of a line and a parabola 
Common points of a line and a parabola we determine by solving their equations as the system of two
equations in two unknowns, 
(1)
y = mx
+ c 
(2)
y^{2} =
2px
(1)
=> (2)
m^{2}x^{2}
+ 2(mc

p)x + c^{2}
=
0, 
therefore,
the coordinates of intersections of a line and a parabola 


Condition for a line to be the tangent to the parabola
 tangency condition

In the formulas for calculating coordinates of intersections there is the expression under the square root
whose value determines three possible cases regarding mutual position of a line and a parabola, 
so for
p >
0 and, 
p

2mc >
0
 the line intersects the parabola at two points S_{1}(x_{1}, y_{1}) and S_{2}(x_{2}, y_{2}), 
p

2mc
= 0  the line is the
tangent of the parabola and have one point of contact D((p

mc)/m^{2},
p/m)) 
or by substituting p
= 2mc, the
tangency point D(c/m,
2c), 
p

2mc
< 0  the
line and the parabola do not intersect. 
If we write the above conditions as 

then 

these
three cases can be explained graphically as
the relation between parameters
m
and c
of the line and the position of the focus F(p/2,
0).

At the yintercept
N
of the line drawn is a
perpendicular which intersects the xaxis at
M,
then

ON
=  c  and OM
=  c · tan
a

=  m · c .

Therefore, when the point
M
is located;

 to the left of
F, i.e.,
if m
· c
< p/2

the line intersects the parabola at
S_{1}and S_{2},

 to the right
of F, i.e.,
if m
· c >
p/2

the line and the parabola do not intersect,

 at the focus F
or if m
· c
= p/2

the line is the tangent of the parabola.





The equation of the tangent and the normal at the point on the parabola

In the equation of the line
y

y_{1}
= m(
x

x_{1})
through the given point we express the slope m
by the given

ordinate of the tangency point,




and since the coordinate of the tangency point must
satisfy the equation of the parabola, then


obtained
is 
y_{1}y
= p(x
+ x_{1}) 
the
equation of


the tangent at the point
P(x_{1}, y_{1})
on the parabola.

Since


the above equation


can be written
using coordinates of the tangency point





As the slope of the normal 

then
the equation of the normal at P(x_{1}, y_{1}),




or 




Parabola and line, examples 
Example:
Find the angle between tangents drawn at intersection points of a line and the parabola
y^{2} = 2px
if the line passes through the focus F(7/4,
0) and its slope
m =
4/3. 
Solution:
As F(p/2,
0)
then, p/2 =
7/4 and the equation
of the parabola y^{2} =
7x.

By plugging
m =
4/3 and
F(7/4,
0) into
the equation of the line
y

y_{1} =
m(x

x_{1})
obtained is


Intersections of the line and the parabola,





Substituting
coordinates of S_{1}
and S_{2}
in


we
get the equation of tangents, 


Slopes of tangents satisfy the perpendicularity
condition, 



Example:
Find the point on the parabola
y^{2} = 9x closest to the line
9x + 4y + 24 =
0. 
Solution:
The tangency point of the tangent parallel to
the given line is the closest point.

9x + 4y +
24 = 0
=> y =
(9/4)x

6, m_{t} =

9/4

The slope of the tangent must satisfy tangency
condition
of the parabola,

p = 2mc
<= m_{t} =

9/4, p
=
9/2

9/2 = 2 · (9/4)
· c
=>
c
= 
1

therefore, the
tangent t
:: y =
(9/4)x

1.

The solution
to the system of equations of the tangent and the parabola
gives the tangency point, that
is







Example:
Given is the polar
4x + y + 12 =
0 of the parabola
y^{2} = 4x, find coordinates of the pole and
write equations of the corresponding tangents. 
Solution:
Intersections of the polar and the parabola are the
tangency points of tangents drawn from the pole P.
Thus, by solving the system of equations of the polar and the parabola
we get the tangency points. 
(1) 4x + y +
12 =
0, (1) =>
(2) (4x

12)^{2} = 4x

(2) y^{2} =
4x
4x^{2}
+ 25x + 36 =
0,



Equation of the tangent at the point on the parabola,





The intersection of tangents is the pole P.
Therefore, we solve the system formed by their equations, 









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