# Parabola (S .N. Dey ) | Ex-4 | Part-6

###### In the previous article, we have solved few Short answer type questions of Parabola Chapter . In this article, we have solved Short answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.

Find the equation of the parabola whose focus is at the origin and the equation of the directrix is

Solution.

Focus Directrix

Let be any point on the parabola.

Also,

For parabola,

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

Solution.

Focus : Directrix :

Let be any point on the parabola.

We know,

Distance between and Perpendicular distance to directrix from

The equation of the directrix of a parabola is and the co-ordinates of its focus are . Find the equation of the parabola.

Focus : Directrix :

Let be any point on the parabola.

We know,

Distance between and Perpendicular distance to directrix from

Find the co-ordinates of vertex and the length of latus rectum of the parabola whose focus is and the directrix is the line

Solution.

We know that the directrix axis of the parabola.

Now, the slope of the directrix

So, the slope of the axis

Equation of the axis which passes through the point and having slope is

Let be the point of intersection of the straight lines and

Solving and we get

Since vertex is the mid-point of and

Hence, the length of the latus rectum

A point moves in such a way that its distance from the point is equal to its distance from the line   Find the equation of its path. What is the name of the curve ?

Solution.

Let at any instant the co-ordinates of the moving point be

The distance between the point and is

Again, the distance of the point  from the from the line is

By question,

Hence, the locus of the point is

.

We know that if a point moves in such a way that it is always equidistant from a fixed point and from a fixed straight line, then the locus of the point will be parabola. So, the name of the curve is parabola.

The co-ordinates of a moving point   are ; show that the locus of is a parabola.

Solution.

From and we get

Hence, by we can say that the locus of the given point is a parabola.

If is a variable parameter, show that the equations represent the equation of a parabola. Find the co-ordinates of vertex, focus and the length of latus rectum of the parabola.

Solution.

We know that

Hence, by we can conclude that the given equations represent a parabola.

Here,

Vertex

Focus

Length of the latus rectum

Find the equation of the circle, one of whose diameters is the latus rectum of Show that this circle goes through the common point of the axis and the directrix of the parabola.

Solution.

For the parabola the length of the latus rectum is unit and the focus is

Since the diameter of the circle is the latus rectum of the parabola, so the centre of the circle is the mid-point of the latus rectum which is

Now, the equation of the circle having centre and with diameter is given by

So, the intersection of the axis and the directrix is

Now putting in the left hand side of we get

So, the circle passes through the point

Hence, we can conclude that this circle goes through the common point of the axis and the directrix of the parabola.