Question 1: Find the equation of the parabola whose:

(i) focus is and the directrix is

(ii) focus is and the directrix is

(iii) focus is and the directrix

(iv) focus is and the directrix

Answer:

(i) Given focus is and the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

(ii) Given focus is and the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

(iii) Given focus is and the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

(iv) Given focus is and the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

Question 2: Find the equation of the parabola whose focus is the point and directrix is the line . Also, find the length of its latus-rectum.

Answer:

Given focus is and the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

Latus Rectum = length of the perpendicular from ( 2, 3) on

Question 3: Find the equation of the parabola, if

(i) the focus is at and the vertex is at

(ii) the focus is at and the vertex is at

(iii) the focus is at and the vertex is at

(iv) the focus is at arid the vertex is at

(v) the focus is at and vertex is at the intersection of the lines and

Answer:

(i) Given focus is at and the vertex is at

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let be the point of intersection of axis and the directrix. The vertex is the midpoint of the line segment joining the focus and the point of intersection of the axis and directrix. Therefore is the midpoint of the line segment joining and .

Therefore,

and

Thus the directrix meets the axis at

Let be the slope of . Then,

Let be the slope of the directrix. Since directrix is perpendicular to .

Thus, the directrix passes through the point and has a slope of . Therefore the equation of the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

(ii) Given focus is at and the vertex is at

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let be the point of intersection of axis and the directrix. The vertex is the midpoint of the line segment joining the focus and the point of intersection of the axis and directrix. Therefore is the midpoint of the line segment joining and .

Thus the directrix meets the axis at

Therefore the equation of directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

(iii) Given focus is at and the vertex is at

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let be the point of intersection of axis and the directrix. The vertex is the midpoint of the line segment joining the focus and the point of intersection of the axis and directrix. Therefore is the midpoint of the line segment joining and .

Thus the directrix meets the axis at

Let be the slope of . Then,

Let be the slope of the directrix. Since directrix is perpendicular to .

Thus, the directrix passes through the point and has a slope of . Therefore the equation of the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

(iv) Given focus is at arid the vertex is at

Thus the directrix meets the axis at

Let be the slope of . Then,

Let be the slope of the directrix. Since directrix is perpendicular to .

Thus, the directrix passes through the point and has a slope of . Therefore the equation of the directrix is

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

(v) Given focus is at and vertex is at the intersection of the lines and

Solving the two equations simultaneously, we vet vertex

Given focus is at and the vertex is at

Thus the directrix meets the axis at

Let be the slope of . Then,

Let be the slope of the directrix. Since directrix is perpendicular to .

Let be any point on the parabola whose focus is and the directrix is .

Draw perpendicular from on the directrix . Then by definition,

This is the equation of the required parabola.

Question 4: Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:

i)

(ii)

iii)

iv)

v)

vi)

vii)

viii)

ix)

Note:

Coordinates of Vertex | ||||

Coordinates of focus | ||||

Equation of the directrix | ||||

Equation of the axis | ||||

Length of Latus rectum | ||||

Focal distance of a point |

Answer:

i)

The given parabola equation is of the form , where

Therefore:

Vertex: The coordinate of vertex are

Focus: The coordinate of vertex are

Axes: The equation of the axes is

Directrix: The equation of the axes is

Latus-rectum: The length of the latus-rectum

(ii)

Therefore:

Vertex: The coordinate of vertex are

Axes: The equation of the axes is

iii) The given equation is

Shifting the origin to the point without rotating the axes and denoting the new coordinates with respect to , we have

Using these relations equation (i), reduces to

This is of the form

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are So the coordinate of the vertex with respect to the old axes are

iv) The given equation is

Shifting the origin to the point without rotating the axes and denoting the new coordinates with respect to , we have

Using these relations equation (i), reduces to

This is of the form

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are So the coordinate of the vertex with respect to the old axes are

v) The given equation is

Shifting the origin to the point without rotating the axes and denoting the new coordinates with respect to , we have

Using these relations equation (i), reduces to

This is of the form

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are So the coordinate of the vertex with respect to the old axes are

vi) The given equation is

Using these relations equation (i), reduces to

This is of the form

Therefore,

vii) The given equation is

Using these relations equation (i), reduces to

This is of the form

Therefore,

viii) The given equation is

Using these relations equation (i), reduces to

This is of the form

Therefore,

ix) The given equation is

Using these relations equation (i), reduces to

This is of the form

Therefore,

Question 5: For the parabola find the extremities of a double ordinate of length . Prove that the lines from the vertex to its extremities are at right angles.

Answer:

Let be the double ordinate of length of the parabola

Then,

Let Then, the coordinates of respectively.

Since lies on

So, coordinates of respectively.

Also, the coordinates of the vertex

Clearly,

Hence is perpendicular to

Therefore the lines from the vertex to the extermities are at right angles.

Question 6: Find the area of the triangle formed by the lines joining the vertex of the parabola to the ends of its latus-rectum.

Answer:

The given equation of the parabola:

This is of the form On comparing we get

Therefore the coordinate of the focus is

Let lie of the parabola. Therefore,

Therefore

Therefore Area of

Question 7: Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is and directrix is . Find also the length of latus-rectum.

Answer:

The equation of directrix is

Slope of directrix

The focus lies on the axis of parabola, therefore

Question 8: At what point of the parabola is the, abscissa three times that of ordinate?

Answer:

Let the ordinate of the required point be

Therefore the abscissa is

Therefore the coordinate of the point is

This point lie on the parabola , hence, will satisfy the equation.

Since

Therefore the required point is

Question 9: Find the equation of a parabola with vertex at the origin, the axis along x-axis and passing through .

Answer:

Case 1:

Case 2:

Either case the equation remains the same which is

Question 10: Find the equation of the parabola with vertex at origin and the directrix .

Answer:

Let be the coordinates of the point of intersection of the axis and the directrix.

Therefore

We know that the vertex is the midpoint of the line joining and focus

Therefore the coordinates of the focus is

By definition of parabola,

Hence the required equation of the parabola is

Question 11: Find the equation of the parabola whose focus is and having vertex at .

Answer:

In a parabola, vertex is the midpoint of the focus and the point of intersection of the axis and directrix.

Let be the the coordinates of point of intersection of the axis and directrix.

Therefore is the midpoint of the line segment joining and

Therefore the directrix meets the axis at

Let be the vertex and be the focus of the required parabola.

Let be the slope of the directrix. Since directrix is perpendicular to the axis,

Thus the directrix passes through with a slope of

Therefore the equation is:

Let be a point on the parabola.

Then distance of P from directrix

Hence the equation of the required parabola is:

Question 12: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wire attached to the cable, the longest wire being 30 m and the shortest wire being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Answer:

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis.

This can be diagrammatically represented as Here, and are the longest and the shortest wires, respectively, attached to the cable.

is the supporting wire attached to the roadway, from the middle.

The equation of the parabola is of the form (as it is opening upwards).

Since is a point on the parabola,

The x-coordinate of point

Therefore

Thus, the length of the supporting wire attached to the roadway from the middle is approximately .

Question 13: Find the equations of the lines joining the vertex of the parabola to the point on it which have abscissa 24.

Answer:

Let and be the points on the parabola and and be the lines joining the vertex to the points and and whose abscissa are

Now

Hence the lines are given by

Question 14: Find the coordinates of points on the parabola whose focal distance is 4.

Answer:

We have

Comparing it with general equation of the parabola we get

Let the required point be

Given Focal distance

Now, the point will satisfy the equation of parabola

Therefore

Hence the coordinate of the points are and

Question 15: Find the length of the line segment joining the vertex of the parabola and a point on the parabola where the line-segment makes an angle to the x-axis

Answer:

Let the coordinates of the point on the parabola be

Let be the line segment.

In the right triangle

Therefore

Now, the curve is passing through the point

Therefore

Question 16: If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Answer:

As the vertex and focus lie on y-axis, therefore y-axis is the axis of the parabola.

If the directrix meets the axis of the parabola at point then

Hence, the equation of the directrix is

Let be any point in the plane of the focus and directrix and let be the perpendicular distance from to the directrix, then lies on parabola if and only if

Question 17: If the line is tangent to the parabola , then find the value of m.

Answer:

We have

Substituting the value of in we get

Since, a tangent touches the curve at a point, the roots of the above equation must be equal.

Therefore