Parabola
Definitition 1Parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which
the plane misses the vertex and
the plane is parallel to the generator
Definition 2Parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which if
α= angle between generator and axis and
β= angle between plane and axis,
and
α=β
then
the section is a parabola,
in which
eccentricity = \(\frac{\cos \beta}{\cos \alpha}\)
here
the eccentricity is the measure of how far the conic deviates from being circular
Definition 3Parabola is a plane curve defined a locus of a point in which
\( \frac{\text{distance from a fixed point}}{\text{distance from a fixed line}} \)= constant (e=1)
In this definition of conic section,
the constant ratio is called eccentricity, it is denoted by e.
the fixed point is called focus.
the fixed line is called directrix.
Parabola is a plane curve defined a locus of a point in which the distance from a fixed point (or focus) and distance from a fixed line (or directrix) is always equal.
Parabola is a plane curve defined a locus of a point which is always equidistant from a fixed point (or focus) to a fixed line (or directrix)
In this definition
 Focus: The fixed point of parabola
 Directrix: The fixed line of parabola
 Axis: The line passing through focus and perpendicular to directrix
 Vertex: The intersecting point of axis and parabola
In a parabola \(y^2=4ax\)
 Focal Distance: Distance between a point on the parabola and the focus.
 Focal Chord: Any chord on the parabola passing through the focus
 Latus Rectum: A chord on the parabola passing through the focus and perpendicular to the axis.
The length of the meeting points of latus rectum to the parabola is called length of latus rectum.
Standard Equation of Parabola
Let C be a parabola whose
Focus is F (a,0)
Directrix is \(l: x = a\)
Vertex is O: (0,0)
Take any point P(x,y) on parabola C,
 Draw PA ⊥ \(l\) then A (a,y)
 Join F and P
By the definition of parabola
PA = PF
or
\(PA^2 = PF^2\)
or
\( (x+a)^2=(xa)^2+y^2 \)
or
\(x^2+2ax+a^2=x^22ax+a^2+y^2\)
or
\(2ax=2ax+y^2 \)
or
\(y^2=4ax \)
Summary on Equation of Parabola
The basic parameters of parabola are summarized as below
Parabola  Parabola  Parabola  Parabola  Parabola 
Equation  \( y^2 =4 a x \)  \( x^2 =4 a y \)  \( (yk)^2 =4 a (xh) \)  \( (xh)^2 =4 a (yk) \) 
Vertex  (0,0)  (0,0)  (h,k)  (h,k) 
Focus  (a,0)  (0,a)  (h+a,k)  (h,k+a) 
Directrix  x=a  y=a  x=ha  y=ka 
Axis Axis of Symmetry 
y=0  x=0  y=k  x=h 
Endpoints of Latus Rectum  (a,±2a)  (±2a,a)  (h+a,k±2a)  (h±2a,k+a) 
Length of Latus Rectum  4a  4a  4a  4a 
The graph of parabola \( x^2=4ay\) is a Ushaped curve . If the parabola opens up, then \(a > 0\). If the parabola opens down, \(a < 0 \). The parabola is also symmetric with a axis line, which is drawn through the vertex, called the axis of symmetry.
Given a parabola \(x^2=4ay\), based on the value of \(a\) , there are two possibilities in the graph. \( a>0\)[a=1]
 \( a<0\) [a=1]
The graph of parabola \( y^2=4ax\) is a Cshaped curve . If the parabola opens right, then \(a > 0\). If the parabola opens left, \(a < 0 \). The parabola is also symmetric with a axis line, which is drawn through the vertex, called the axis of symmetry.
Given a parabola \(y^2=4ax\), based on the value of \(a\) , there are two possibilities in the graph. \( a>0\)[a=1]
 \( a<0\) [a=1]
Parametric equation of Parabola
A parametric equation of a parabola describes the coordinates of points on the parabola using a parameter, usually denoted as 𝑡. For a standard parabola, \(y^2=4ax\), the parametric equations is\(x=at^2, y=2at\)
General equation of Parabola
The general equation of a parabola can be derived from its geometric definition. A parabola is the set of all points in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix.Let the focus be \( F(h, k) \) and the directrix be the line \( ax+by+c=0\).
Then, the distance of a point \( (x, y) \) and the focus \( (h, k) \) is
\( \sqrt{(x  h)^2 + (y  k)^2} \)
Also, the distance from a point \( (x, y) \) to the line \(ax+by+c=0 \) is
\( \frac{ax+by+c}{\sqrt{a^2 + b^2}} \)
We know that
Setting these two distances equal , we get
\( \sqrt{(x  h)^2 + (y  k)^2} = \frac{ax+by+ c}{\sqrt{a^2 + b^2}} \)
Squaring both sides , we get
\( (x  h)^2 + (y  k)^2 = \frac{(ax+by+ c)^2}{a^2 + b^2}\)
This is the required equation.
Exercise
 Find the equation of parabola
 vertex at (0,0) focus at (4,0)
Given that
vertex at (0,0)
focus at (4,0) so \((a,0)=(4,0)\)
As we know, the equation of parabola with its vertex at the origin (0,0) and its focus at (a,0) is
\(y^2=4ax\)
Given that the focus is at (4,0), we have \(a = 4\).
Therefore, the equation of the parabola is
\(y^2=4ax\)
or \(y^2=4.4x\)
or \(y^2=16x\)
This completes the solution.
 vertex at (1,3) focus at (5,3)
Given that
vertex at (1,3) so \((h,k)=(1,3)\)
focus at (5,3) so \((h+a,k)=(5,3)\)
Since the axis of parabola is parallel to xaxis [because \(y_1=y_2=3\) in both coordinates of vertex (1,3) and focus (5,3)], So, focul distance from the vertex is
\(h+a=5\)
\((1)+a=5\)
\(a=6\)
Now, the equation of horizontally opening parabola with its vertex at \((h,k)=(1,3)\) and focul distance a=6 is
\((yk)^2=4a(xh)\)
or \((y3)^2=4.6(x+1)\)
or \((y3)^2=24(x+1)\)
This completes the solution.
 vertex at (2,3) focus at (2,5)
Given that
vertex at (2,3) so \((h,k)=(2,3)\)
focus at (2,5) so \((h,a+k)=(2,5)\)
Since the axis of the parabola is parallel to the yaxis [because \(x_1=x_2=2\) in both coordinates of the vertex (2,3) and focus (2,5)], the focal distance from the vertex is
\( a+k=5\)
\( a+(3)=5\)
\( a=2\)
Now, the equation of vertically opening parabola with its vertex at \((h,k)=(2,3)\) and focal distance \(a=2\) is
\((x  h)^2 = 4a(y  k)\)
or \((x  2)^2 = 4 \cdot 2 \cdot (y  3)\)
or \((x  2)^2 = 8(y  3)\)
This completes the solution.
 vertex at (1,2) focus at (0,2)
Given that
vertex at (1,2) so \((h,k)=(1,2)\)
focus at (0,2) so \((h+a,k)=(0,2)\)
Since the axis of parabola is parallel to xaxis [because \(y_1=y_2=3\) in both coordinates of vertex (1,2) and focus (0,2)], So, focul distance from the vertex is
\( h+a=0\)
\( (1)+a=5\)
\( a=1\)
Hizontally opening parabola with its vertex at \((h,k)=(1,2)\) and focul distance \(a=1\) is
\((yk)^2=4a(xh)\)
or \((y2)^2=4.(1)(x1)\)
or \((y2)^2=4(x1)\)
This completes the solution.
 vertex at (5,3) and end of the latus rectum (1,5) and (1,11)
Given that
vertex at (5,3) so \((h,k)=(5,3)\)
End of latus rectum are (1,5) and (1,11)
The ends of the latus rectum have the same xcoordinate (1). This means the parabola opens horizontally (left or right).
Next, the focus of the parabola is
\(\left ( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
\(\left ( \frac{(1)+(1)}{2}, \frac{5+(11)}{2} \right) \)
\(( 1,3 ) \) so \((h+a,k)=( 1,3)\)
Since the axis of parabola is parallel to xaxis [because \(y_1=y_2=3\) in both coordinates of vertex (5,3) and focus (1,3)], So, focul distance from the vertex is
\( h+a=1\)
\( (5)+a=1\)
\( a=4\)
Now, the equation of horizontally opening parabola with its vertex at (h,k) and focul distance a is
\((yk)^2=4a(xh)\)
or \((y+3)^2=4.(4)(x+5)\)
or \((y+3)^2=16(x+5)\)
This completes the solution.
 point of intersection of directrix and axis is at (0,4) and focus at (6,4)
Given that
focus is (6,4) so \((h+a,k)=(6,4)\)
Next, the vertex of the parabola is, mid point of (0,4) and (6,4), which is
\(\left ( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
\(\left ( \frac{0+6}{2}, \frac{4+4}{2} \right) \)
\(( 3,4 ) \) so \((h,k)=(3,4 )\)
Since the axis of parabola is parallel to xaxis [because \(y_1=y_2=3\) in both coordinates of vertex (3,4) and focus (6,4)], So, focul distance from the vertex is
\( h+a=6\)
\( (3)+a=6\)
\( a=3\)
Now, the equation of horizontally opening parabola with its vertex at (h,k) and focul distance a is
\((yk)^2=4a(xh)\)
or \((y4)^2=4.(3)(x3)\)
or \((y4)^2=12(x3)\)
This completes the solution.
 vertex at (0,0) focus at (4,0)
 Find the equation of parabola
 vertex at (1,2) directrix at x=4
Given that
vertex is (1,2) so \((h,k)=(1,2)\)
directrix is \(x = 4\) so \(x = ha\)
Thus, we have
\(x = ha\)
\(4 = (1)a\)
\(a=5\)
Now, the equation of the horizontally opening parabola with its vertex at (h,k) and focal distance a is
\((yk)^2=4a(xh)\)
or \((y2)^2=4 \cdot (5) \cdot (x+1)\)
or \((y2)^2=20(x+1)\)
This completes the solution.
 vertex at (1,1) directrix at y=3
Given that
vertex is (1,1) so \((h,k)=(1,1)\)
directrix is \(y = 3\) so \(y=ka\)
Thus, we have
\(y=ka\)
\(3 = (1)a\)
\(a=2\)
Now, the equation of the vertically opening parabola with its vertex at (h,k) and focal distance a is
\((x  h)^2 = 4a(y  k)\)
or \((x + 1)^2 = 4 \cdot (2) \cdot (y  1)\)
or \((x + 1)^2 = 8(y  1)\)
This completes the solution.
 focus at (1,2) directrix at x=3
Given that
focus is (1,2) so so \((h+a,k)=(1,2)\)
directrix is \(x = 3\) so \(x=ha\)
So, focul distance from the vertex is
\( x=ha\)
\( 3=h+a2a\)
\( 3=(1)2a\)
\( a=2\)
Also,
\( h+a=1\)
\( h+(2)=1\)
\( h=1\)
Now, the equation of the horizontally opening parabola with its vertex at \((h,k)=(1,2)\) and focal distance \(a=2\) is
\((y  k)^2 = 4a(x  h)\)
or \((y  2)^2 = 4 \cdot 2 \cdot (x + 1)\)
or \((y  2)^2 = 8(x + 1)\)
This completes the solution.
 focus at (1,2) directrix at y=4
Given that
focus is (1,2) so \((h,k+a)=(1,2) \)
directrix is \(y = 4\) so \(y=ka\)
So, focul distance from the vertex is
\( y=ka\)
\( (4)=k+a2a\)
\( (4)=(2)2a\)
\( a=1\)
Also
\( k+a=2\)
\( k+(1)=2\)
\( k=3\)
Now, the equation of the vertically opening parabola with its vertex at \((h,k)=(1,3)\) and focal distance \(a=1\) is
\((x  h)^2 = 4a(y  k)\)
or \((x  1)^2 = 4 \cdot 1 \cdot (y + 3)\)
or \((x  1)^2 = 4(y + 3)\)
This completes the solution.
 focus at (3,4) directrix at 2xy+5=0
Given that
focus is (3,4)
directrix is \(2x  y + 5 = 0\)
Let (x,y) be the arbitrary poinon the parabola, then
distance from directrix=distance from focus
We know that the distance from directrix \(Ax + By + C = 0\) is given by
\(d_1 = \frac{Ax + By + C}{\sqrt{A^2 + B^2}}\)
or \(d_1 = \frac{2x  y + 5 }{\sqrt{2^2 + 1^2}}\)
or \(d_1 = \frac{2x  y + 5 }{\sqrt{5}}\)
Next, we know that the distance from focus (3,4) is given by
\(d_2 = \sqrt{(x+3)^2+(y4)^2} \)
Now, the equation of parabola is obtained by
\(d_1^2=d_2^2 \)
or\(\frac{(2x  y + 5 )^2}{5}=(x+3)^2+(y4)^2 \)
This completes the solution.
 vertex at (1,2) directrix at x=4
 Given below are the equation of parabola. Find the coordimtes of the focus, the vertex, the equation of the directrix, the length of the latus rectum, and the equation of the axis.
 \(y^2=16x\)
Given the equation of the parabola:
\(y^2 = 16x\)
BYy comparing the given equation with the standard form of the parabola, we get
Standard form: \(y^2 = 4ax\)
By comparing, we get that
\(4a = 16 \implies a = \frac{16}{4} = 4\)
Now

The vertex of the parabola in the standard form \(y^2 = 4ax\) is at the origin:
Vertex: \((0,0)\)

The focus of the parabola is at \((a,0)\):
Focus: \((4,0)\)

The directrix of the parabola is the vertical line \(x = a\):
Directrix: \(x = 4\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot 4 = 16\)

The axis of the parabola is the horizontal line passing through the vertex and parallel to the xaxis:
Axis: \(y = 0\)

The vertex of the parabola in the standard form \(y^2 = 4ax\) is at the origin:
 \(x^2=12y\)
Given the equation of the parabola:
\(x^2 = 12y\)
By comparing the given equation with the standard form of the parabola, we get
Standard form: \(x^2 = 4ay\)
By comparing, we get that
\(4a = 12 \implies a = \frac{12}{4} = 3\)
Now

The vertex of the parabola in the standard form \(x^2 = 4ay\) is at the origin:
Vertex: \((0,0)\)

The focus of the parabola is at \((0, a)\):
Focus: \((0, 3)\)

The directrix of the parabola is the horizontal line \(y = a\):
Directrix: \(y = 3\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot 3 = 12\)

The axis of the parabola is the vertical line passing through the vertex and parallel to the yaxis:
Axis: \(x = 0\)

The vertex of the parabola in the standard form \(x^2 = 4ay\) is at the origin:
 \((y2)^2=4x12\)
Given the equation of the parabola:
\((y2)^2 = 4x  12\)
By comparing the given equation with the standard form of the parabola, we get
Standard form: \((y  k)^2 = 4a(x  h)\)
By comparing, we get that
\((y  2)^2 = 4(x  3)\)
\(4a = 4 \implies a = 1\)
Now

The vertex of the parabola in the standard form \((y  k)^2 = 4a(x  h)\) is at \((h, k)\):
Vertex: \((3, 2)\)

The focus of the parabola is at \((h + a, k)\):
Focus: \((3 + 1, 2) = (4, 2)\)

The directrix of the parabola is the vertical line \(x = h  a\):
Directrix: \(x = 3  1 = 2\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot 1 = 4\)

The axis of the parabola is the horizontal line passing through the vertex and parallel to the xaxis:
Axis: \(y = 2\)

The vertex of the parabola in the standard form \((y  k)^2 = 4a(x  h)\) is at \((h, k)\):
 \((x+1)^2+8y16=0\)
Given the equation of the parabola:
\((x+1)^2 + 8y  16 = 0\)
By comparing the given equation with the standard form of the parabola, we get
Standard form: \((x  h)^2 = 4a(y  k)\)
Rearranging the given equation to match the standard form:
\((x + 1)^2 = 8y + 16\)
\((x + 1)^2 = 8(y  2)\)
By comparing, we get that
\(4a = 8 \implies a = 2\)
Now

The vertex of the parabola in the standard form \((x  h)^2 = 4a(y  k)\) is at \((h, k)\):
Vertex: \((1, 2)\)

The focus of the parabola is at \((h, k + a)\):
Focus: \((1, 2  2) = (1, 0)\)

The directrix of the parabola is the horizontal line \(y = k  a\):
Directrix: \(y = 2 + 2 = 4\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot 2 = 8\)

The axis of the parabola is the vertical line passing through the vertex and parallel to the yaxis:
Axis: \(x = 1\)

The vertex of the parabola in the standard form \((x  h)^2 = 4a(y  k)\) is at \((h, k)\):
 \(y^2+4x+8=0\)
Given the equation of the parabola:
\(y^2 + 4x + 8 = 0\)
By comparing the given equation with the standard form of the parabola, we get
Standard form: \(y^2 = 4a(x  h)\)
Rearranging the given equation to match the standard form:
\(y^2 = 4x  8\)
\(y^2 = 4(x + 2)\)
By comparing, we get that
\(4a = 4 \implies a = 1\)
Now

The vertex of the parabola in the standard form \(y^2 = 4a(x  h)\) is at \((h, k)\):
Vertex: \((2, 0)\)

The focus of the parabola is at \((h + a, k)\):
Focus: \((2  1, 0) = (3, 0)\)

The directrix of the parabola is the vertical line \(x = h  a\):
Directrix: \(x = 2 + 1 = 1\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot 1 = 4\)

The axis of the parabola is the horizontal line passing through the vertex and parallel to the xaxis:
Axis: \(y = 0\)

The vertex of the parabola in the standard form \(y^2 = 4a(x  h)\) is at \((h, k)\):
 \(x^2=2y+6\)
Given the equation of the parabola:
\(x^2 = 2y + 6\)
By comparing the given equation with the standard form of the parabola, we get
Standard form: \(x^2 = 4a(y  k)\)
Rearranging the given equation to match the standard form:
\(x^2 = 2(y + 3)\)
\(x^2 = 2(y  (3))\)
By comparing, we get that
\(4a = 2 \implies a = \frac{2}{4} = \frac{1}{2}\)
Now

The vertex of the parabola in the standard form \(x^2 = 4a(y  k)\) is at \((h, k)\):
Vertex: \((0, 3)\)

The focus of the parabola is at \((0, k + a)\):
Focus: \((0, 3 + \frac{1}{2}) = (0, \frac{5}{2})\)

The directrix of the parabola is the horizontal line \(y = k  a\):
Directrix: \(y = 3  \frac{1}{2} = \frac{7}{2}\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot \frac{1}{2} = 2\)

The axis of the parabola is the vertical line passing through the vertex and parallel to the yaxis:
Axis: \(x = 0\)

The vertex of the parabola in the standard form \(x^2 = 4a(y  k)\) is at \((h, k)\):
 \(y^2=6y12x+45\)
Given the equation of the parabola:
\(y^2 = 6y  12x + 45\)
By comparing the given equation with the standard form of the parabola, we get
Standard form: \((y  k)^2 = 4a(x  h)\)
Rearranging the given equation to match the standard form:
\(y^2  6y = 12x + 45\)
To complete the square for \(y\), we add and subtract \(\left(\frac{6}{2}\right)^2 = 9\):
\(y^2  6y + 9 = 12x + 45 + 9\)
\((y  3)^2 = 12x + 54\)
\((y  3)^2 = 12(x  4.5)\)
By comparing, we get that
\(4a = 12 \implies a = 3\)
Now

The vertex of the parabola in the standard form \((y  k)^2 = 4a(x  h)\) is at \((h, k)\):
Vertex: \((4.5, 3)\)

The focus of the parabola is at \((h + a, k)\):
Focus: \((4.5  3, 3) = (1.5, 3)\)

The directrix of the parabola is the vertical line \(x = h  a\):
Directrix: \(x = 4.5 + 3 = 7.5\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot 3 = 12\)

The axis of the parabola is the horizontal line passing through the vertex and parallel to the xaxis:
Axis: \(y = 3\)

The vertex of the parabola in the standard form \((y  k)^2 = 4a(x  h)\) is at \((h, k)\):
 \(x^2=4x3y+5\)
Given the equation of the parabola:
\(x^2 = 4x  3y + 5\)
By comparing the given equation with the standard form of the parabola, we get
Standard form: \((x  h)^2 = 4a(y  k)\)
Rearranging the given equation to match the standard form:
\(x^2  4x = 3y + 5\)
To complete the square for \(x\), we add and subtract \(\left(\frac{4}{2}\right)^2 = 4\):
\(x^2  4x + 4 = 3y + 5 + 4\)
\((x  2)^2 = 3y + 9\)
\((x  2)^2 = 3(y  3)\)
By comparing, we get that
\(4a = 3 \implies a = \frac{3}{4}\)
Now

The vertex of the parabola in the standard form \((x  h)^2 = 4a(y  k)\) is at \((h, k)\):
Vertex: \((2, 3)\)

The focus of the parabola is at \((h, k + a)\):
Focus: \((2, 3  \frac{3}{4}) = (2, \frac{9}{4})\)

The directrix of the parabola is the horizontal line \(y = k  a\):
Directrix: \(y = 3 + \frac{3}{4} = \frac{15}{4}\)

The length of the latus rectum is \(4a\):
Length of the latus rectum: \(4 \cdot \frac{3}{4} = 3\)

The axis of the parabola is the vertical line passing through the vertex and parallel to the yaxis:
Axis: \(x = 2\)

The vertex of the parabola in the standard form \((x  h)^2 = 4a(y  k)\) is at \((h, k)\):
 \(y^2=16x\)
 Find the point on the prabola \(y^2=8x\) at which the ordinate is double the abscissa
Given the equation of the parabola:
\(y^2 = 8x\)
We need to find the point on the parabola where the ordinate (y) is double the abscissa (x). Let the coordinates of the point be \((x, y)\). Thus, we have:
\(y = 2x\)
Substitute \(y = 2x\) into the equation of the parabola:
\((2x)^2 = 8x\)
\(4x^2 = 8x\)
\(4x^2  8x = 0\)
\(4x(x  2) = 0\)
Solving this equation, we get:
\(x = 0\) or \(x = 2\)
For \(x = 0\), we have:
\(y = 2 \cdot 0 = 0\)
Point: \((0, 0)\)
For \(x = 2\), we have:
\(y = 2 \cdot 2 = 4\)
Point: \((2, 4)\)
Therefore, the points on the parabola \(y^2 = 8x\) where the ordinate is double the abscissa are:
\((0, 0)\) and \((2, 4)\)
This completes the solution.
 Find the point on the prabola \(x^2=12y\) at which the abscissa is twice the ordinate
Given the equation of the parabola:
\(x^2 = 12y\)
We need to find the point on the parabola where the abscissa (x) is twice the ordinate (y). Let the coordinates of the point be \((x, y)\). Thus, we have:
\(x = 2y\)
Substitute \(x = 2y\) into the equation of the parabola:
\((2y)^2 = 12y\)
\(4y^2 = 12y\)
\(4y^2  12y = 0\)
\(4y(y  3) = 0\)
Solving this equation, we get:
\(y = 0\) or \(y = 3\)
For \(y = 0\), we have:
\(x = 2 \cdot 0 = 0\)
Point: \((0, 0)\)
For \(y = 3\), we have:
\(x = 2 \cdot 3 = 6\)
Point: \((6, 3)\)
Therefore, the points on the parabola \(x^2 = 12y\) where the abscissa is twice the ordinate are:
\((0, 0)\) and \((6, 3)\)
This completes the solution.
 Find the point on the prabola \(y^2=8x\) at which the ordinate is double the abscissa
 Find the equation of parabola with vertex at the origin, axis parallel to xaxis, and passing through the point (1,3)
Given that the vertex of the parabola is at the origin and the axis is parallel to the xaxis, the standard form of the parabola is:
\((y  k)^2 = 4a(x  h)\)
Since the vertex is at the origin \((0, 0)\), the equation simplifies to:
\(y^2 = 4ax\)
We are given that the parabola passes through the point \((1, 3)\). Substituting \(x = 1\) and \(y = 3\) into the equation:
\((3)^2 = 4a \cdot 1\)
\(9 = 4a\)
\(a = \frac{9}{4}\)
Therefore, the equation of the parabola is:
\(y^2 = 4 \cdot \frac{9}{4} \cdot x\)
\(y^2 = 9x\)
This completes the solution.
 Find the equation of parabola with focul width 16, axis parallel to xaxis, and passing through the points (3,7) and (3,1)
\((y  k)^2 = 4a(x  h)\)
The focal width (length of the latus rectum) is given by \(4h\). Hence:
\(4a = \pm 16\)
\(a = \frac{16}{4} = \pm 4\)
The points (3,7) and (3,1) lie on the parabola, the vertex of the parabola is at the midpoint of the xcoordinates, so
\(k=\frac{7 + (1)}{2} = 3\)
So, the vertex is
Vertex: \((h,k)=(h,3)\)
Now, the required equation of parabola is
\((y  3)^2 = 4 a (x  h)\)
\((y  3)^2 = \pm 16 (x  h)\)
Now, substitute a point, say (3,7) into the standard form of the parabola to get
\((y  3)^2 = \pm 16 (x  h)\)
\((7  3)^2 = \pm 16 (3  h)\)
\(4^2 = \pm 16 (3h)\)
\(1 = \pm (3h)\)
\(h=2, a=4\) or \(h=4, a=4\)
Therefore, the equation of the parabola is:
\((y  3)^2 = 16 (x  2)\) or \((y  3)^2 =  16 (x  4)\)
This completes the solution.
 What is the equation of the parabola whose length of latus recutus is 16, axis along the line parallel to xaxis passing through points (3,3) and (3,2)?
\((y  k)^2 = 4a(x  h)\)
The focal width (length of the latus rectum) is given by \(4h\). Hence:
\(4a = \pm 16\)
\(a = \frac{16}{4} = \pm 4\)
The points (3,3) and (3,2) lie on the parabola, the vertex of the parabola is at the midpoint of the xcoordinates, so
\(k=\frac{3 + (2)}{2} = 0.5\)
So, the vertex is
Vertex: \((h,k)=(h,0.5)\)
Now, the required equation of parabola is
\((y  3)^2 = 4 a (x  h)\)
\((y  0.5)^2 = \pm 16 (x  h)\)
Now, substitute a point, say (3,3) into the standard form of the parabola to get
\((y  0.5)^2 = \pm 16 (x  h)\)
\((3  0.5)^2 = \pm 16 (3  h)\)
\(6.25 = \pm 16 (3h)\)
\( \pm \frac{6.25}{16} = (3h)\)
\(h=3\frac{6.25}{16}, a=4\) or \(h=3+\frac{6.25}{16}, a=4\)
\(h=\frac{41.75}{16}, a=4\) or \(h=\frac{54.25}{16}, a=4\)
Therefore, the equation of the parabola is:
\((y  0.5)^2 = 16 (x  \frac{41.75}{16})\) or \((y  0.5)^2 =16 (x  \frac{54.25}{16})\)
This completes the solution.
 Find the equation of parabola with axis parallel to yaxis, length of latus rectum 8 and passing through the points (2,1) and (2,1)
\((x  h)^2 = 4a(y  k)\)
The length of the latus rectum is given by \(4a\). Hence:
\(4a = \pm 8\)
\(a = \frac{8}{4} = \pm 2\)
The points \((2, 1)\) and \((2, 1)\) lie on the parabola, the vertex of the parabola is at the midpoint of the xcoordinates, so
\(h=\frac{2 + (2)}{2} = 0\)
So, the vertex is
Vertex: \((h,k)=(0,k)\)
Now, the required equation of parabola is
\(x^2 = 4a(y  k)\)
Now, substitute a point, say (2,1) into the standard form of the parabola to get
\(x^2 = 4a(y  k)\)
\(x^2 =\pm 8(y  k)\)
\(2^2 = \pm 8(1  k)\)
\(0.5 = \pm (1  k)\)
\(k=0.5, a=2\) or \(k=1.5, a=2\)
Therefore, the equation of the parabola is:
\(x^2 = 8(y  0.5)\) or \(x^2 = 8(y  1.5)\)
This completes the solution.
 Find the equation of parabola with vertex at the origin, axis parallel to xaxis, and passing through the point (1,3)
 The equation of the parabola is \(y^2=16x\). Find the coordinate of the ends of the latus rectum. Also find the area of the triangle formed by joining the ends the the latus rectum and the vertex of the parabola.
Given the equation of the parabola:
\(y^2 = 16x\)
By comparing the given equation with the standard form of the parabola, we get:
Standard form: \(y^2 = 4ax\)
By comparing, we get that:
\(4a = 16 \implies a = \frac{16}{4} = 4\)
Now:

The vertex of the parabola is at \((0, 0)\).

The focus of the parabola is at \((a, 0) = (4, 0)\).

The length of the latus rectum is \(4a = 16\). The ends of the latus rectum are at (a,2a) and (a,2a), which is
\((4, 8)\) and \((4, 8)\)

To find the area of the triangle formed by joining the ends of the latus rectum and the vertex of the parabola, use the vertices \((0, 0)\), \((4, 8)\), and \((4, 8)\). The area of the triangle can be calculated using the formula:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(= \frac{1}{2} 16.4\)
\(= 32\)
\((4, 8)\) and \((4, 8)\)
The area of the triangle formed by joining these points with the vertex is:
32
This completes the solution.

The vertex of the parabola is at \((0, 0)\).
 A double ordinate of the parabola is \(y^2=2ax\) is of length 4a. Prove that the line joining the vertex to its ends are at right angles.
Given that, double ordinate of the parabola is \(y^2=2ax\) is of length 4a, so
ends of the double ordinate are \((2a, 2a)\) and \((2a, 2a)\)
The vertex of the parabola is at \((0, 0)\). To prove that the lines joining the vertex to the ends of the double ordinate are perpendicular, we calculate the slopes of these lines:
For the line joining the vertex \((0, 0)\) and the point \((a, 2a)\):
Slope m1= \(\frac{2a  0}{2a  0} = 1\)
For the line joining the vertex \((0, 0)\) and the point \((2a, 2a)\):
Slope = \(\frac{2a  0}{2a  0} = 1\)
The product of the slopes of two lines that are perpendicular is \(1\). Here:
Product = \(1 \cdot (1) = 1\)
Hence, the lines joining the vertex to the ends of the double ordinate are at right angles.
This completes the proof.
 Find the equation of the circle circumscribing the segment of the parabola \(y^2=8x\) cut off by the latus rectum of the parabola
Given the parabola \(y^2 = 8x\), comparing with \(y^2 = 4ax\), we get
\(4a = 8 \implies a = 2\)
Thus, the coordinates of the ends of the latus rectum are \((a, \pm 2a)\) which is
\((2, 4)\) and \((2, 4)\)
Given the points \((2, 4)\), \((2, 4)\), we need to find the equation of the circle that circumscribes the points
(2, 4), (2, 4) and vertex(0,0)
Here, the center of the circle lies on xaxis, so is \((h, 0)\).
The radius of the circle is distance from center to any of the points (2, 4), (2, 4) and (0,0), so is [distance from (0,0)]
\(h\)
So, the standard form of a circle with center \((h, 0)\) and radius \(h\) is:
\((x  h)^2 + y^2 = h^2\)
\(x^2 + y^2 2hx=0\)
Putting th epoint (2,4), we get
\(2^2 + 4^2 2h.2=0\)
\(h=5\)
So, the requiredequation of the circle, is
\(x^2 + y^2 10x=0\)
This completes the solution.
 Find the equation of the circle circumscribing the segment of the parabola \(y^2=4ax\) cut off by the latus rectum of the parabola
Given the parabola \(y^2 = 4ax\), the coordinates of the ends of the latus rectum are \((a, \pm 2a)\) which is
\((a, 2a)\) and \((a, 2a)\)
Given the points \((a, 2a)\) and \((a, 2a)\), we need to find the equation of the circle that circumscribes the points
(a, 2a), (a, 2a) and vertex(0,0)
Here, the center of the circle lies on xaxis, so is \((h, 0)\).
The radius of the circle is distance from center to any of the points (a, 2a), (a, 2a) and (0,0), so is [distance from (0,0)]
\(h\)
So, the standard form of a circle with center \((h, 0)\) and radius \(h\) is:
\((x  h)^2 + y^2 = h^2\)
\(x^2 + y^2 2hx=0\)
Putting the point (a,2a), we get
\(a^2 + (2a)^2 2h.(a)=0\)
\(h=\frac{5a}{2}\)
So, the required equation of the circle, is
\(x^2 + y^2 5ax=0\)
This completes the solution.
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