









Parabola and line intersection




Check if a point is inside of a parabola





Notice that this parabola has the vertex at the origin and focus distance of p from the directrix.
If the vertex is located at (h , k) then the parabola equation becomes:
Notes:



Type  Sketch  parabola equation  Focus at Vertex 
Directrix Symmetry 


1  x^{2} = 2py 



2  x^{2} = −2py 



3  y^{2} = 2px 



4  y^{2} = −2px 




Type  sketch  parabola equation  Focus at Vertex 
Directrix Symmetry 


1  (x − a)^{2} = 2p(y − b) 



2  (x − a)^{2} = −2p(y − b) 



3  (y − b)^{2} = 2p(x − a) 



4  (y − b)^{2} = −2p(x − a) 


















We can see that the signs of A and C must be positive (equals to square values).



Given the parabola x^{2} + 2y − 3x + 5 = 0, find the vertex, focus, directrix and the axis of symmetry.


By the method of completing the square we write the equation of the parabola as:
From the last equation we notice that a = 3/2 b = −11/8 and p = 1



Given the parabola y^{2} −4y + 2x + 7 = 0, find the vertex, focus, directrix and axis of symmetry.


We will solve the general solution of the parabola equation: y^{2} + Ay + Bx + C = 0
By the method of completing the square we write the equation of the parabola as:
From the parabola given values we have: A = −4 B = 2 C = 7
From the last equation we see that the coordinate of the vertex is:
The sign of B is negative so the parabola open side is to the left direction.



Find the coordinate of the foci and the directrix equation of the parabola given by the equation
y^{2} = −12x. 




Given the parabola x^{2} − 2x − 2y + 3 = 0, find the vertex, focus, directrix and axis of symmetry.





Given the directrix line of a parabola by the equation y = 1.75 and the focus at the point
F(−2 , 0.25). Find the equation of the parabola and the vertex and the symmetry line equations. 




Given the focus of a parabola at (1 , 4) and the directrix equation x + y − 9 = 0 find the equation of the parabola and the coordinates of (x_{d} , y_{d}).





The focus of a parabola is located at the point (0 , 2) and the vertex at (0 , 4) find the
equation of the parabola. 




The vertex of a parabola is located at the point (−1.5 , 2) and the directrix equation is
x = −1 find the coordinate of the vertex and the equation of the parabola. 




Find the transformation equations from the polynomial form x^{2} + ax + by + c = 0 to the standard form of the parabola (x + h)^{2} = p(y + k) and vice verse.


Now we will change the parenthesis value into square expression
From the last expression we can see that
p is twice the distance from the focus to the directrix line p = b / 2
The open side of the parabola is pointing down if b is negative and pointing up if b is positive.



The focus of a parabola is located at the point (−1.5 , 1) and the vertex at (−1 , 1) find the
equation of the parabola the equation of the directrix and the equation of the symmetry line. 

The directrix line is vertical and is passing through the point (−0.5 , 1) x = − 0.5
And the symmetry line equation from the sketch is: y = 1



Find the general expression for the intersection of a parabola x^{2} + Ax + By + C = 0
and the line y = mx + q
With the expression developed find the intersection of the following parabola and line
x^{2} + 2x − 4y + 3 = 0 and the line x + 3y − 5.5 = 0. 

And the intersection points are (−4.33 , 3.28) and (1 , 1.5)



Find the intersection points of a parabola (y − 1)^{2} − 2(x + 2.5) = 0 and the lines a) x = 0 and b) the line y = 0.


And the intersection points are (0 , −1.24) and (0 , 3.24)
b) Substitute the line value y = 0 to the parabola equation we get: −2x − 4 = 0
It can be seen that there is only one solution for x and it is at x = −2 therefore only one intersection point exists at (−2 , 0)



Find the intersection points of the a parabola x^{2} − 2xy + y^{2} + 10x + 22y − 71= 0 and the lines a) x + y − 1 = 0. b) x = 4. c) y = 0. Also derive a general expression for the intersection points.


a) Substitute the value of y from the line equation y = − x + 1 into the parabola equation.
x^{2} − 2x (− x + 1) + (− x + 1)^{2} + 10x + 22( − x + 1) − 71 = 0
b) Substitute x = 4 into the parabola equation: 16 − 8y + y^{2} + 40 + 22y −71 = 0
y^{2} + 14y − 15 = 0
c) Substitute y = 0 into the parabola equation: x^{2} + 10x −71 = 0
The general expression for the intersection points of:
Substitute the value of y into the equation of the parabola to get the values of x:
We got a quadratic equation and the solutions are the values of x. The values of y can be found according to equation (2) .
Substitute the value of x into the equation of the parabola to get the values of x:
We got a quadratic equation and the solutions are the values of y. The values of x can be found according to equation (2) .



Find the intersection points of the parabola (y − 2)^{2} = 4(x − 1) and the lines
2x + y − 8 = 0. Also find the tangent lines at the intersection points. 

Line equation when a point and the slope are given is: y = mx + y_{1} − mx_{1}



Find the general equation of the line perpendicular to the parabola given by the equation
(x + h)^{2} = p(y + k) Then use the equation to find the perprndicular line to the parabola (x − 4)^{2} = −3(y + 2) at the point (1 , −5). 

If the tangency point is at (x_{1} , y_{1}) then the equation of the perpendicular line is:



A parabola is given by y^{2} −2y −2x − 4 = 0, find, a) the equation of the tangent line at a point on the parabola where y = 4 and b) the tangent lines when x = −2.


By applying the implicit derivation on the parabola, we get the slope of the tangent line:
b) To find the y value of the point x = −2 we must solve the equation y^{2} − 2y −2x − 4 = 0
The solutions are y_{1} = 0 and y_{2} = 2 and the two tangent points are: (−2 , 0) and (−2 , 2)



Find the equations of the tangent lines to the parabola given by the equation (y + 1)^{2} = x − 5
at the point where x = 6. 

First we find the corresponding y values at x = 6: (y + 1)^{2} = 6 − 5
Solving this equation, we get: y^{2} + 2y = 0 and the solutions are y_{1} = 0 and y_{2} = −2
Now applying the implicit derivation on the parabola, to get the slope dy/dx of the tangent line:
Substitute the values of y into equation (1) in order to find the slopes of the tangent lines:



Find the equations of the tangent lines to the tilted parabola given by the equation
x^{2} + 4xy + 4y^{2} − 2x + 12y + 8 = 0 and the intersection points of the line x = 0 and the parabola. 

First step is to find the coordinates of the line and parabola so we shell substitute the value of the line x = 0 into the parabola equation y^{2} + 3y + 2 = 0
We found the two intersection points (0 , −1) and (0 , −2) and we have to find the tangent lines at those points.
To find the slope of the tangent line we shell explicitly derivate the equation of the parabola



Find the equation and the vertex point of a parabola. If the focus point is at (−1 , −2) and directrix at the line x − 2y + 3 = 0.


The vertex is located half way between the focus and the point G.
Point G is the intersection of the symmetry line and the directrix.
The symmetry line is passing through the focus and is perpendicular to the directrix line.



Find the equation of a parabola with focus at point (4 , 1) and vertex at point (−2 , 3) .


According to the parabola definition the distance EG = FE



Find the equation of a parabola with focus at point (−1 , 2) and the intersection point of the symmetry line and the directrix at (3 , 4) .


According to the parabola definition the distance EG = FE



Given a parabola with focus at point (2 , − 1) and vertex at point (4 , 1) determined if points
(1 , 3) and point (5 , − 2) are inside or outside the parabola. 

And the coefficients are: A = 1 B = 1 C = −9
Now we have to find the relations between the lengths FB and BE



Given the parabola x^{2} − 2xy + y^{2} + 10x + 6y − 87 = 0 determined if the points (4 , 6) and the
point (6 , 1) are inside or outside of the parabola. 

According to example 5b the value of q from eq (9) is:
The directrix equation according to slope and point (d_{x} , d_{y}) is: y = mx + (y_{1} − mx_{1})
According to the parabola definition we have: FP = PA (where P is the point to verify).



Given the parabola y^{2} − 4y − 4x + 8 = 0 determined if the points (1 , 0) and (2 , 0) and (3 , 0)
are inside or outside of the parabola. 

According to equations (14) and (15) the value of the intersection of symmetry axis and the directrix is (6 , 7)
The directrix equation is passing through point (0,2) so the equation of the directrix is x = 0
According to the parabola definition we have: FP = PA (where P is the point to verify).
