NCERT Class 11-Math՚S: Exemplar Chapter – 11 Conic Sections Part 3 (For CBSE, ICSE, IAS, NET, NRA 2022)

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11.2 Solved Examples

Short Answer Type

Question 1:

Find the centre and radius of the circle


We write the given equation in the form

Now, completing the squares, we get

Comparing it with the standard form of the equation of the circle, we see that the centre of the circle is and radius is .

Question 2:

If the equation of the parabola is , find coordinates of the focus, the equation of the directrix and length of latus rectum.


The given equation is of the form ay where a is positive.

Therefore, the focus is on y-axis in the negative direction and parabola opens downwards.

Comparing the given equation with standard form, we get .

Therefore, the coordinates of the focus are and the equation of directrix is and the length of the latus rectum is a, i.e.. , .

Question 3:

Given the ellipse with equation , find the major and minor axes, eccentricity, foci and vertices.


We put the equation in standard form by dividing by and get

This shows that and . Hence Since the denominator of is larger, the major axis is along , minor axis along , foci are and

and vertices are and .

Question 4:

Find the equation of the ellipse with foci at and as one of the directories.


We have which give . Therefore,

Now . Thus, the equation of the ellipse is

Question 5:

For the hyperbola , find the vertices, foci and eccentricity.


The equation of the hyperbola can be written as so and , so that , which gives . Vertices are and foci are .

Question 6:

Find the equation of the hyperbola with vertices at and . Find its foci.


Since the vertices are on the y-axes (with origin at the mid-point) , the equation is of the form .

As vertices are , so the required equation of the hyperbola is and the foci are .

Long Answer Type

Question 7:

Find the equation of the circle which passes through the points , and . Find its centre and radius.


By substitution of coordinates in the general equation of the circle given by , we have

From these three equations, we get


Hence, the equation of the circle is

Therefore, the centre of the circle is and radius is .