# Parabola (S .N. Dey ) | Ex-4 | Part-8

###### In the previous article, we have solved few Short answer type questions (11-15) of Parabola Chapter . In this article, we will solve Very Short answer type questions of Parabola Chapter (Ex-4) of S.N.Dey mathematics, Class 11.

Find the focus , the length of the latus rectum and the directrix of the parabola

Solution.

The given parabola can be written as

Comparing the  parabola with we get,

Focus :

Length of the latus rectum

The directrix of the parabola :

Find the length of the latus rectum of the parabola

Solution.

The given equation of the parabola can be written as

Hence, the length of the latus rectum of the parabola is

Find the equation of the parabola whose co-ordinates of the vertex and focus are and  respectively.

Solution.

The given parabola has the vertex and the focus

Clearly, the axis of the parabola is parallel to the axis.

Hence, the equation of the parabola is

The vertex of a parabola is at the origin and its focus is find the equation of the parabola.

Solution.

The given parabola has the vertex and the focus

Clearly, the axis of the parabola is parallel to the axis.

Hence, the equation of the parabola is

The parabola passes through the point find the co-ordinates of focus and the length of latus rectum.

Solution.

Since the parabola passes through the point we have,

So, the equation of the parabola turns out to be

Comparing the parabola with the parabola where , we get the focus and the length of the latus rectum is given by

The parabola goes through the point of intersection of and Find its focus.

Solution.

From and , we get

From we get,

Hence, the point of intersection of and is

Since the parabola passes through the point

So, the equation of the parabola is

Hence, the focus of the parabola is

A parabola having a vertex at the origin and axis along axis passes through find the equation of the parabola.

Solution.

The equation of the parabola having vertex at and axis along axis can be written as

Since the parabola is passing through the point , so we get by

Hence, the equation of the parabola can be written as

Find the equation of the parabola whose vertex is and directrix is the line

Solution.

The equation of the directrix is

So, the equation of the parabola is along the positive axis and can be written as

Here, the distance between the directrix and the vertex of the parabola unit.

Find the equation of the parabola whose vertex is at the origin and directrix is the line

Solution.

The equation of the directrix is

So, the equation of the directrix is along the negative axis and can be written as where the distance between the directrix and the vertex of the parabola unit.

If the parabola passes through the point of intersection of the straight lines and find the co-ordinates of the focus and the length of its latus rectum.

Solution.

From , we get

From

So, the point of intersection of the given straight lines is

Since the parabola passes through the point

So, the co-ordinates of its focus

The length of its latus rectum is

The parabola passes through the centre of the circle find the co-ordinates  of the focus , length of the latus rectum and the equation of the directrix.

Solution.

The equation of the given circle can be written as

Comparing with , we get

The centre of the given circle is

Since the given parabola passes through the point ,

So, the focus of the parabola

The length of the latus rectum is

The equation of the directrix is given by

If the parabola passes through the centre of the circle what is the length of the latus rectum of the parabola.

Solution.

The equation of the given circle can be written as

Comparing with , we get

The centre of the given circle is

Since the given parabola passes through the point ,

The length of the latus rectum is

Find the point on the parabola at which the ordinate is double the abscissa.

Solution.

Let be a point on the parabola

By question,

So,

Since the point lies on the given parabola,

Hence,

The point on the parabola at which the ordinate is three times the abscissa.

Solution.

Let be a point on the parabola

By question,

So,

Since the point lies on the given parabola,

Find the point on the parabola which forms a triangle of area with the vertex and focus of the parabola.

Solution.

Let be a point on the parabola . Also, suppose that the vertex and the focus

So, area of the triangle

What type of conic is the locus of the moving point Find the equation of the locus.

Solution.

Let

Hence, by and we get,

Hence, by we can conclude that the equation of the conic represents a parabola.

The focal distance of a point on the parabola is find the co-ordinates of the point.

Solution.

Let the coordinate of that point is

Equation of the parabola

Now, the co-ordinates of focus

By question,

But since can not be  negative, so

So,

So, the coordinates are

Find the focal distance of a point on the parabola if the abscissa of the point be

Solution.

The given equation of the parabola is

So, the focus

If the abscissa of the point on the given parabola is then we need to find the ordinate of that point.

Hence, the focal distance = distance between and

Determine the positions of the points and with respect to the parabola

Solution.

From we can conclude that first point lies outside the parabola, second point lies inside the parabola and third point lies on the given parabola.

For what values of  will the point be an inside point of the parabola

Solution.

The given equation of the parabola can be written as We know if lies inside the parabola, then

So, if the point lies  inside the given parabola, then