## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 11 Conic Sections

**Short Answer** **Type Questions **

**(Score 3)**

Question 1.

a. Equation represents a vertical ellipse if

a. a^{2} = b^{2 }b. a^{2}>b^{2}

c.a^{2}<b^{2}

d.a>b

b. Find the equation of the circle passing through the point (2,4) and having centre at the intersection of the lines x – y = 4 and 2x + 3y + 7 = 0.

Answer:

a. a^{2} < b^{2 }b. Solving x – y = 4 and 2x + 3y + 7 = 0

we get x= 1,y=-3

Centre(1,-3) Circle passes through (2,4)

Equation of circles is

(x -1)^{2} + (y + 3)^{2} = 50

x^{2} + y^{2} – 2x + 6y – 40 = 0

Question 2.

a. For the parabola x^{2}=-4 ay, the focus is at

a. (0,0)

b. (a, 0)

c. (-a, 0)

d.(0,-a)

b. What is the radius of the circle (x + 2)^{2 }+ y^{2} = 25 ?

Answer:

a. (0, -a)

x^{2}=-4ay

Equation of axis is y-axis Company with equation x^{2} = 4ay; a= -a

focus of the parabola (0, -a)

b. General equation of a circle with centre (h,k) and radius r is (x-h)^{2} + (y – k)^{2} = r^{2 }given (x + 2)^{2} + y^{2} =25

(x -2)^{2} + (y – 0 )^{2} = 5^{2}; r = 5

Question 3.

If the origin is one end of the diameter of a circle whose centre is (-1,2), then find the other end of the diameter.

Answer:

Centre of circle C (-1,2)

Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) be the end of diameter. C is the midpoint of AB, then

Question 4.

9x^{2} – 16y^{2} = 144 is a hyperbola, with electricity =……………..

a. 5

b. 6

c. 7

d. 3

ii. The area of a circle with diameter 8 is greater than the area of an ellipse with 2a=8. Check whether this statement is true or false.

Answer:

Question 5.

Find the equation of the ellipse with centre at origin, length of major axis is

5 and length of minor axis is 4.

Answer:

Genaral equation of ellipse is

Question 6.

a. Circles x^{2} + y^{2} + 6x + 6y = 0 and x^{2} + y^{2}– 12x- 12y = 0

a. touch each other internally

b. touch each other externally

c. intersect in two points

d. cut orthogonally

b. y^{2} = 8x represents a parabola, find its directrix.

Answer:

b. y^{2} = 8x . The given equation involves y^{2}, so the axis of symmetry is along the x- axis. The coefficient of x is positive so the parabola opens to the right comparing with

y^{2} = 4ax; y^{2} = 8x = 4×2×;c; a = 2

Directrix of the parabola is x = -2;x + 2 = 0

Question 7.

Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ±5).

Answer:

If the foci are on y axis, thew major axis is along the y axis. So equation of the ellipse is of the them.

Question 8.

Find the equation of the ellipse, if foci are (+3,0 ) and a = 4.

Answer:

Given, foci are on X-axis. So, the major axis will be along the X-axis. So, the equation of ellipse is form

Question 9.

Find the equation of the ellipse having length of minor axis = 16 and foci (0, +6). h Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.

Answer:

a. Length of minor axis = 2b = 16 ⇒ b = 8

Foci are (0, + 6) ⇒ c = 6

a^{2} =b^{2} + c^{2} = 8^{2} + 6^{2} = 64 + 36 = 100

Major axis on Y-axis. Then, equation of ellipse is

b. Let the centre of circle in first quadrant be (a,a).

Equation of circle is (x-a)^{2} + (y – a)^{2} = a^{2 }x^{2} + a^{2} -2ax + y^{2} + a^{2} – 2ay

= a^{2 }x^{2} + y^{2} – 2ax – 2ay + a^{2} = 0

Question 10.

Find the equation of a circle with centre (3, -2) and touching the X-axis.

Answer:

Here, centre of a circle (3, -2) with radius is 2 units.

**Short Answer** **Type Questions **

**(Score 4)**

Question 1.

a. Find the focus of the parabola x^{2} – 8x + 2y + 7 = 0.

b “The graph of is an ellipse”. State whether this statement is true or false.

c. Find the equation of the ellipse with foci (+3,0) and a = 4.

Answer:

Question 2.

Find the equation of the ellipse satisfying following conditions.

i. Foci at (± 5, 0) and x = 36/5 as one directrix.

ii. Find length of latus rectum and vertices of the ellipse.

Answer:

Question 3.

Match the following

Answer:

i. The parabola is open upwards and hence the equation is of the form x^{2} = 4ay.

Hence the parabola is x^{2} = 12y

ii. The parabola is open leftwards hence the equation is of the form y^{2} = – 4ax.

Hence the parabola is y^{2} = -12x.

iv. The hyperbola is 16x^{2} – 9y^{2} = 144

Question 4.

i.“It is possible for a parabola to intersect its directrix”. Check whether this statement is true or false?

ii. Find the equation of ellipse with the following conditions:

a. Foci at (± 3,0) and passing through (4,1)

b. Foci at (± 4,0) and vertices at (± 5,0)

Answer:

Question 5.

The line x -1 =0 is the directrix of a parabola y^{2} = kx, then

a. Find the value of k.

b. Find the vertex, focus, axis of parabola and length of latus rectum of the parabola.

Answer:

a. k=-4

b. Vertex = (0,0) Focus = (-1,0)

Length of latus rectum = 4a = 4×1=4 Axis of parabola is x axis.

Question 6.

In the figure S ans S’ are foci of the ellipse,

a Find the co-ordinates of foci,

b. Find the distance between S and S’.

c. What is the maximum area of the triangle PSS’?

Answer:

a. foci = (+3,0)

b. length of SS’ = 6

c. Maximum Area = ^{1}/2 × 6 × 4= 12

Question 7.

Find the equation of ellipse with the following conditions:

i. Passing through (4, 3) and (-1, 4) and having axes along coordinate axes.

ii. e = 3/4, foci on y-axis, centre at origin and passing through (6,4)

Answer:

i. Let ………..(1)

be the ellipse

Since (1) passes through (4, 3) we get Similarly using (-1,4) we get

……………….(2)

similarly using (-1,4) we get

**Long Answer** **Type Questions **

**(Score 6)**

Question 1.

Complete the table given below:

Answer:

i.x^{2} = -16y 4a = -16⇒a = -4

Focus: (0, 4)

Equation of directrix: y = 4

Vertex:(0,0) Latusrectum=4a=|-16|=16

ii. x^{2} = 6y ∴4a = 6⇒a = 3/2 Focus: (0,3/2)

Equation of directrix: y = -3/2 Vertex: (0,0) Latus rectum = 4a = 6

iii. y^{2} = 8x 4a = 8 ⇒ a = 2 Focus: (2,0) Directrix: x=-2,

Vertex: (0,0) Latus rectum = 4a = 8

iv. y^{2} = -12x 4a = -12 ⇒ a = -3 Focus: (3,0) Directrix: x=3 Vertex: (0,0)

Latus rectum = 4a = -121 = 12

v. y^{2} = – 6x4a = -6 ⇒ a = -3/2

Focus: (3/2,0) Directrix: x = 3/2,

Vertex: (0,0)

Latus rectum = 4a = |-61 |= 6

Question 2.

Complete the table given below:

Answer:

Question 3.

a. Find the equation of ellipse with foci at (0, ± 1) and length of Latus rectum 15/2. h It is possible for an ellipse to have the foci outside the ellipse. Check whether this statement is true or false?

Answer:

Question 4.

a. The circle whose equation is x^{2} + (y – 1)^{2}= 2 has the centre………………….

b. Consider the equation

Answer:

a.{0,1)

b. i. Ellipse

ii. b^{2} = a^{2}(1 -e^{2})

25= 100(1 -e^{2}); (1 – e^{2}) = 1/4

i.e, e = 3/4 or e =

Question 5.

a. “If the vertex and focus of a parabola are on a horizontal line, then the directrix of the parabola is vertical”. State whether this statement is true or false,

b. 9x^{2} – 16y^{2} = 144 is a hyperbola with essentricity………….

Answer:

a. True

b. 9x^{2}-16y^{2}= 144

Question 6.

i. Consider the equation x^{2}=-4ay. Its graph is …………

ii. Find the coordinates of the focus and length of the latus rectum of the parabola y^{2}=-6x

iii. The equation of a hyperbola is y^{2} – 16x^{2 }= 16. Write the equation in standard form, hence find the centre, foci vertices, eccentricity and length of latus rectum.

Answer:

Question 7.

a. Find the centre and radius of the circle x^{2} + y^{2} -2x + 2y – 23 = 0.

b. Find the equation of the circle concentric with x^{2}+y^{2}-2x+2y – 23=0 and radius

c. Find the equation of the circle passing through (1, -1) and the centre at the intersection of the lines x – y = 4 and 2x+3y = -7.

Answer:

a. x^{2} – 2x + y^{2} + 2y = 23

(x^{2} – 2x + 1) + (y^{2} + 2y + 1) = 25

(x- 1)^{2} + (y+1)^{2} = 25 Centre (1,-1)

Radius = = 5

b. Equation of a circle with centre (1,-1) and radius is

Question 8.

The two end points of a diameter of a circle are (-2,3) and (3, -5).

i. Determine the centre and radius of the circle.

ii. Determine equation of the circle.

iii. Determine whether the origin lies inside the circle or not.

Answer:

Distance between origin and centre is less than that of radius. Therefore origin is inside the circle.

**NCERT Questions and Answers**

Question 1.

Find an equation of the circle with centre at (0,0) and radius r.

Answer:

Here h = k = 0. Therefore, the equation of the circle is x^{2} + y^{2} = r^{2}

Question 2.

Find the centre and the radius of the circle

x^{2} + y^{2} + 8x + 10y – 8 = 0

Answer:

The given equation is (x^{2} + 8x) + (y^{2} +1 Oy) = 8 Now, completing the squares within the parenthesis,

we get

(x^{2} +8x+ 16) + (y^{2}+ 10y + 25) = 8+ 16 + 25

i.e. (x + 4)^{2} + (y + 5)^{2} = 49

i.e. (x-(-4)}^{2} + {y — (—5)}^{2} = 7^{2 }Therefore, the given circle has centre at (- 4, -5) and radius 7

Question 3.

Find the equation of the circle which passes through the points (2, – 2), and (3,4) and whose centre lies on the line x + y = 2.

Answer:

Let the equation of the circle be (x – h)^{2} + (y-k)^{2} = r^{2}.

Since the circle passes through (2, – 2) and (3,4), we have

(2 – h)^{2}+ (-2 – k)^{2} = r^{2}…………… (1)

and (3 – h)^{2} + (4 – k)^{2} =r^{2}…………… (2)

Also since the centre lies on the line x + y = 2,

we have h + k = 2……………. (3)

Solving the equations (1), (2) and (3), we get h = 0.7,k= 1.3 and r^{2} = 12.58 Hence, the equation of the required circle is (x-0.7)^{2} + (y-1.3)^{2} =12.58.

Question 4.

Find the centre and radius of the circle 2x^{2} +2y^{2} – x = 0

Answer:

Question 5.

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through (2,3).

Answer:

Let (b, 0) be the centre (a point on x-axis) radius, r = 5

Equation of circle is (x – h)^{2} + (y – 0)^{2} = 5^{2 }(x – h)^{2} + y^{2}= 25………………. (1)

(2,3) is a point on (1)

(2 -h)^{2} + 3^{2} = 25, (2 -h)^{2}= 16,

2 -h = ±4, h = -2 or h = 6

Put h = -2 in(1) ⇒(x + 2)^{2} + y^{2}= 25

i.e, x^{2} + y^{2} + 4x-21 =0

Put h = 6in(1) ⇒(x-6)^{2} + y^{2}= 25

i.e, x^{2} + y^{2}– 12x+ 11=0

Question 6.

Find the equation of the circle passing through the points (2,3) and (-1,1) and whose centre is on the line x – 3y – 11 = 0

Answer:

Let the equation be (x – h)^{2}+(y – k)^{2} = r^{2 }Circle passes through (2, 3) ⇒

(2-h)^{2}+(3-k)^{2} = r^{2}……………………. (1)

Circle passes through (-1, 1) ⇒

(-1-h)^{2}+(1 -k)^{2} = r^{2}……………………. (2)

Centre is on x – 3y – 11 = 0 ⇒

h-3k-11=0…………………… (3)

Solving (1), (2) and (3)

x^{2} + y^{2}-7x + 5y-14 = 0

Question 7.

Find die equation of the parabola with vertex at (0,0) and focus at (0,2).

ans:

Since the vertex is at (0,0) and the focus is at (0,2) which lies on y-axis, the y-axis is the axis of the parabola. Therefore, equation of the parabola is of the form x^{2} = 4 a y. Thus, we have x^{2} = 4 (2) y,

i.e.,x^{2} =8y

i. Focus

ii. Equation of directrix

iii. Length of latus rectum

Answer:

4a = 12 ⇒a = 3

i. Focus is (a, 0); i.e, (3,0)

ii. Directrix is x = -a, i.e, x = -3

iii. Length of latus rectum = 4a = 12

Question 8.

A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point Alies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP=6 cm. Show that the locus of P is an ellipse.

Answer:

Let AB be the rod making an angle 9 with OX and P (x, y) the point on it such that AP = 6cm.

Since AB = 15 cm, we have PB = 9 cm.

From P draw PQ and PR perpendiculars on y- axis and x-axis, respectively.

Question 9.

A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point Alies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP=6 cm. Show that the locus of P is an ellipse.

Answer:

Let AB be the rod making an angle 9 with OX and P (x, y) the point on it such that AP = 6cm.

Since AB = 15 cm, we have PB = 9 cm.

From P draw PQ and PR perpendiculars on y- axis and x-axis, respectively.