# Parabola (S .N. Dey ) | Ex-4 | Part-1

In this article, we have solved 6 Long answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.

Find the equation of the parabola whose vertex is and the equation of latus rectum is Find the co-ordinates of the point of intersection of this parabola with its latus rectum.

Solution.

The equation of the straight line passing through the point and perpendicular to is

So, the co-ordinates of the focus of the parabola are

So, the required equation of parabola is

Now, the co-ordinates of the point of intersection of the parabola with its latus rectum are given by and

The co-ordinates of the two ends of latus rectum of a parabola are and find the equation of the parabola.

Solution.

Since the ordinates of the extremities of the latus rectum are equal, so the latus rectum is parallel to axis and the axis of the parabola is parallel to the axis.

So, the equation of latus rectum is and co-ordinates of focus are given by the mid-point of and which are

Let the axis of the parabola be which passes through So, the axis of the parabola is

The co-ordinates of the vertex of parabola can be and another possible co-ordinates of the vertex of the parabola can be

So, the equation of the parabola with vertex is

Again, the equation of the parabola with vertex is

Find the equation of the parabola whose vertex is and focus is

Solution.

The equation of the axis of the parabola is the straight line joining the points and and is given by

The equation of the directrix which is perpendicular to the straight line is

Now, the co-ordinates of the focus the co-ordinates of the vertex

Let the point of intersection of the axis of the parabola and the directrix be

Since is the mid-point of and , so

Since the straight line passes through

Hence by we get the equation of the directrix

Let be any point on the parabola.

Since lies on the required parabola, hence we must have

Hence, represents the required equation of parabola.

The directrix of a parabola is and vertex is the point Find the position of focus and the equation of the parabola. [Council Sample Question ‘]

Solution.

The equation of the directrix of the parabola is

The axis of the parabola is perpendicular to the directrix and is passing through the vertex

The equation of any straight line perpendicular to can be written as

Since the straight line is passing through  the point ,

So, by   we get the equation of the axis of the parabola which is

Solving and , we get

So, the point of intersection of the axis of parabola and the directrix is

Let be the focus of the parabola and be the mid-point of and

So,

Let be any point on the parabola and is the perpendicular distance from upon the directrix.

The equation represents the required parabola.

The axis of a parabola is parallel to axis and it passes through the points find its equation.

Solution.

Since the axis of the parabola is parallel to axis, the equation of the parabola can be written as

Since the parabola is passing through

Now, the parabola is passing through the points and

From and we get,

Hence, the equation of the required parabola is

A parabola passes through the points and its axis is parallel to axis. Find its equation.

Solution.

By the question, since the axis of the parabola is parallel to axis , the equation of the parabola can be written as

Since the parabola passes through the point ,

So, equation can be written as

Since the parabola passes through the points and

From and we get

So,

Hence, the equation of the parabola is

If the extremities of a focal chord of the parabola be and , prove that

Solution.

The equation of the focal chord joining the points and is

Since the focal chord passes through the point

If be the co-ordinates of an extremity of a focal chord of the parabola the show that the length of the chord is

Solution.

If be the co-ordinates of an extremity of a focal chord of the parabola, another extremity is

So, the length of the chord is

Show that the equation of the chord of the parabola through the points and on it is

Solution.

Since the points and lie on the parabola so

From and we get,

The equation of the straight line joiningÂ  the points and isÂ