EXERCISE 11.1

1. Draw a line segment of length 7.6 cm and divide it in the ratio 5: 8. Measure the two parts.

Answer:

Given: A line segment of length 7.6 cm.

To construct: To divide it in the ration 5: 8 and to measure the two parts.

Steps of construction:

chapter 11-Constructions Exercise 11.1/image001.jpg

Steps of Construction :

(i)     Draw AB = 5.6 cm

(ii)    At a draw an acute ∠BAX below base AB.

(iii)    On AX make 5 + 8 i.e. 13 equal parts and mark them as A

chapter 11-Constructions Exercise 11.1

On measuring, we get

AC = 3.1 cm,

CB = 4.5 cm,

Justification :

chapter 11-Constructions Exercise 11.1

[Using basic proportionally theorem]

chapter 11-Constructions Exercise 11.1

This shows that C divides AB in the ratio 5 : 8.

2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Answer:

To construct: To construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3of the corresponding sides of the first triangle.

Steps of construction:

Step 1

Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.

Step 2

Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

Step 3

chapter 11-Constructions Exercise 11.

Step 4

chapter 11-Constructions Exercise 11.1.

Step 5

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1

Justification:

The construction can be justified by proving that

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1

This justifies the construction.

3. Construct a triangle with sides 6 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.

Answer:

To construct: To construct a triangle of sides 5 cm, 6 cm and 7 cm and then a triangle similar to it whose sides are 7/5 of the corresponding sides of the first triangle.

Steps of construction:

Step 1

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 2

Draw a ray AX making acute angle with line AB on the opposite side of vertex C.

Step 3

chapter 11-Constructions Exercise 11.1

Step 4

chapter 11-Constructions Exercise 11.1

Step 5

chapter 11-Constructions Exercise 11.1.

chapter 11-Constructions Exercise 11.1

Justification:

The construction can be justified by proving that

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle.

Answer:

To construct: To construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then a triangle similar to it whose sides are 3/2 of the corresponding sides of the first triangle.

Steps of construction:

Step 1. Draw a line segment BC = 8 cm.

Step 2. Draw the perpendicular bisector XY of BC, cutting BC at D.

Step 3. With D as centre and radius 4 cm, draw an arc cutting XY at A.

Step 4. Join AB and AC. Here, ∆ABC is an isosceles whose base is 8 cm and altitude is 4 cm.

Step 5. Below BC, draw an acute angle ∠CBX.

Step 6. Along BX, mark three pointschapter 11-Constructions Exercise 11.1/image049.jpg

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1.

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 600 . Then construct a triangle whose sides are 3/4 of the corresponding sides of triangle ABC.

Answer:

A ΔA’BC’ whose sides arechapter 11-Constructions Exercise 11.1of the corresponding sides of ΔABC can be drawn as follows:

Steps of construction:

Step 1

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 3 and 4),chapter 11-Constructions Exercise 11.1

Step 4

chapter 11-Constructions Exercise 11.1

Step 5

chapter 11-Constructions Exercise 11.1.

chapter 11-Constructions Exercise 11.1

6. Draw a triangle ABC with side BC = 7 cm,∠B =450,∠A =1050.Then construct a triangle whose sides are 4/3 times the corresponding sides of ∆ABC.

Answer:

To construct: To construct a triangle ABC with side BC = 7 cm∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3of the corresponding sides of the first triangle ABC.

∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

The required triangle can be drawn as follows.

Steps of construction:

Step 1

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 4 and 3)chapter 11-Constructions Exercise 11.1

Step  4

chapter 11-Constructions Exercise 11.1

Step  5

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1

Justification:

The construction can be justified by proving that

chapter 11-Constructions Exercise 11.1

n ΔABC and ΔA’BC’,

∠ABC = ∠A’BC’ (Common)

∠ACB = ∠A’C’B (Corresponding angles)

∴ ΔABC ∼ ΔA’BC’ (AA similarity criterion)

chapter 11-Constructions Exercise 11.1

On comparing equations (1) and (2), we obtain

chapter 11-Constructions Exercise 11.1

This justifies the construction.

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Answer:

To construct: To construct a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm and then a triangle similar to it whose sides are 5/3 of the corresponding sides of the first triangle ABC.

It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other.

The required triangle can be drawn as follows.

Steps of construction:

Step 1

Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.

Step 2

Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.

Step 3

Draw a ray AX making an acute angle with AB, opposite to vertex C.

Step 4

chapter 11-Constructions Exercise 11.1

Step 5

chapter 11-Constructions Exercise 11.1

Step 6

chapter 11-Constructions Exercise 11.1

chapter 11-Constructions Exercise 11.1

Justification:

The construction can be justified by proving that

chapter 11-Constructions Exercise 11.1

∠ABC = ∠AB’C’ (Corresponding angles)

∠BAC = ∠B’AC’ (Common)

∴ ΔABC ∼ ΔAB’C’ (AA similarity criterion)

chapter 11-Constructions Exercise 11.1

On comparing equations (1) and (2), we obtain

chapter 11-Constructions Exercise 11.1

This justifies the construction.

Exercise 11.2

1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Answer:

Given: A circle whose centre is O and radius is 6 cm and a point P is 10 cm away from its centre.

To construct: To construct the pair of tangents to the circle and measure their lengths.

NCERT solutions for class 10 maths Exercise 11.2/image001.png

Steps of Construction:

(a) Join PO and bisect it. Let M be the mid-point of PO.

(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.

(c) Join PQ and PR.

Then PQ and PR are the required two tangents.

By measurement, PQ = PR = 8 cm

Justification:

Join OQ and OR.

∵ ∠OQP and ∠ORP are the angles in semi circles.

∠OQP = 90° = ∠ORP

Also, since OQ, OR are radii of the circle, PQ and PR will be the tangents to the circle at Q and R respectively.

2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Answer:

To construct: To construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its lengths. Also to verify the measurements by actual calculation.

Steps of Construction:

1. With ‘O’ as centre two circles are constructed with radii 4 cm and 6 cm.

2. Draw OM radius for small circle and drawn perpendicular at M which meets big circle at P and Q.

3. Now PQ is the tangnet drawn for small circle.

NCERT solutions for class 10 maths Constructions/image005.jpg

Tangent PMQ = 9 cm

Verification : In ⊥∆OMP, ∠M = 90°.

chapter 11/image006.png

Similarly, MQ = 4.5 cm.

∴ PQ = PM + MQ = 4.5 + 4.5

∴ PQ = 9 cm.

3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Answer:

To construct: A circle of radius 3 cm and take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre and then draw tangents to the circle from these two points P and Q.

Steps of Construction:

chapter 11-Constructions Exercise 11.2/image013.jpg

1. Draw a line segment PQ of 14 cm.

2. Now, mark the midpoint O of PQ.

3. Draw the perpendicular bisectors of PO and OQ which intersects at points R and S on PQ.

4. With centre R and radius RP draw a circle.

5. With centre S and radius, SQ draw a circle.

6. And now, with centre O and radius 3 cm draw another circle which intersects the previous circles at the points A, B, C, and D.

7. Finally, join PA, PB, QC and QD. Thus, PA, PB, QC, and QD are the required tangents.

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 600.

Answer:

To construct: A pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 600.

Steps of Construction:

chapter 11-Constructions Exercise 11.2/image015.jpg

Step I: Take a point O on the plane of the paper and draw a circle of radius OA = 5 cm.

Step II: Produce OA to B such that OA = AB = 5 cm.

Step III: Taking A as the centre draw a circle of radius AO = AB = 5 cm.

Suppose it cuts the circle drawn in step I at P and Q.

Step IV: Join BP and BQ to get the desired tangents.

Justification: In OAP, we have

OA = OP = 5 cm (= Radius) Also,

AP = 5 cm (= Radius of circle with centre A)

∴ ∆OAP is equilateral.

⇒ ∠PAO = 60º ⇒ ∠BAP = 120º

In ∆BAP, we have

BA = AP and ∠BAP = 120º

∴ ∠ABP = ∠APB = 30º ⇒ ∠PBQ = 60º

5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer:

To construct: A line segment of length 8 cm and taking A as centre, to draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Also, to construct tangents to each circle from the centre to the other circle.

Steps of Construction:

The tangents can be constructed on the given circles as follows

Step 1

Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius.

Step 2

Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AR. These are the required tangents.

NCERT solutions for class 10 maths chapter 11-Constructions Exercise 11.2/image022.jpg

Justification:

The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.

NCERT solutions for class 10 maths /image002.png

∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠ASB = 90°

⇒ BS ⊥ AS

Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.

6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and  ∠B = 900. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer:

To construct: A right triangle ABC with AB = 6 cm, BC = 8 cm and ∠B = 900 .BD is the perpendicular from B on AC and the tangents from A to this circle.

Follow the given steps to construct the figure.

Steps of Construction:

Step 1

Draw a line BC of 8 cm length.

Step 2

Draw BX perpendicular to BC.

Step 3

Mark an arc at the distance of 6 cm on BX. Mark it as A.

Step 4

Join A and C. Thus, ∆ABC is the required triangle.

Step 5

With B as the centre, draw an arc on AC.

Step 6

Draw the bisector of this arc and join it with B. Thus, BD is perpendicular to AC.

Step 7

Now, draw the perpendicular bisector of BD and CD. Take the point of intersection as O.

Step 8

With O as the centre and OB as the radius, draw a circle passing through points B, C and D.

Step 9

Join A and O and bisect it. Let P be the midpoint of AO.

Step 10

Taking P as the centre and PO as its radius, draw a circle which will intersect the circle at point B and G. Join A and G.

Here, AB and AG are the required tangents to the circle from A.

chapter 11-Constructions Exercise 11.2/image024.png

Justification:

The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.

Then, NCERT solutions for class 10 maths /image002.png

∠AGE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠AGE = 90°

⇒ EG ⊥ AG

Since EG is the radius of the circle, AG has to be a tangent of the circle.

Already, ∠B = 90°

⇒ AB ⊥ BE

Since BE is the radius of the circle, AB has to be a tangent of the circle.

7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Answer:

To construct: A circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Steps of Construction:

chapter 11/image025.jpg

The required tangents can be constructed on the given circle as follows.

Step 1

Draw a circle with the help of a bangle.

Step 2

Take a point P outside this circle and take two chords QR and ST.

Step 3

Draw perpendicular bisectors of these chords. Let them intersect each other at point O.

Step 4

Join PO and bisect it. Let U be the mid-point of PO. Taking U as centre, draw a circle of radius OU, which will intersect the circle at V and W. Join PV and PW.

PV and PW are the required tangents.

Justification:

The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to be proved that O is the centre of the circle. Let us join OV and OW

Then, NCERT solutions for class 10 maths chapter 11-Constructions Exercise 11.2/image002.png

We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. ∠PVO is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.

∴ ∠PVO = 90°

⇒ OV ⊥ PV

Since OV is the radius of the circle, PV has to be a tangent of the circle. Similarly, PW is a tangent of the circle.

EXERCISE 11.3

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

 

Question 1. chapter 11-Conic Sections Exercise 11.3/image001.png

Solution :

The given equation ischapter 11-Conic Sections Exercise 11.3/image001.png.

Here, the denominator of ncert solutionis greater than the denominator ofncert solution.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withncert solution, we obtain = 6 and b = 4.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (6, 0) and (–6, 0).

Length of major axis = 2a = 12

Length of minor axis = 2b = 8

ncert solution

Length of latus rectum ncert solution

Question2. chapter 11-Conic Sections Exercise 11.3/image019.png

Solution :

The given equation isncert solution.

Here, the denominator of ncert solutionis greater than the denominator ofncert solution.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withncert solution, we obtain = 2 and a = 5.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (0, 5) and (0, –5)

Length of major axis = 2a = 10

Length of minor axis = 2b = 4

ncert solution

Length of latus rectum ncert solution

Question 3. chapter 11-Conic Sections Exercise 11.3/image031.png

Solution :

The given equation isncert solution.

Here, the denominator of  ncert solution is greater than the denominator of ncert solution.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withncert solution, we obtain = 4 and b = 3.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices arencert solution.

Length of major axis = 2a = 8

Length of minor axis = 2b = 6

ncert solution

Length of latus rectum ncert solution

Question4. chapter 11-Conic Sections Exercise 11.3/image041.png

Solution :

The given equation is ncert solution.

Here, the denominator of ncert solution is greater than the denominator of ncert solution.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withncert solution, we obtain = 5 and a = 10.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (0, ±10).

Length of major axis = 2a = 20

Length of minor axis = 2b = 10

ncert solution

Length of latus rectum ncert solution

Question 5. chapter 11-Conic Sections Exercise 11.3/image051.png

Solution :

The given equation isncert solution.

Here, the denominator of ncert solutionis greater than the denominator ofncert solution.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withncert solution, we obtain = 7 and b = 6.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (± 7, 0).

Length of major axis = 2a = 14

Length of minor axis = 2b = 12

ncert solution

Length of latus rectumncert solution

Question6. chapter 11-Conic Sections Exercise 11.3/image060.png

Solution :

The given equation isncert solution.

Here, the denominator of ncert solutionis greater than the denominator ofncert solution.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withncert solution, we obtain = 10 and a = 20.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (0, ±20)

Length of major axis = 2a = 40

Length of minor axis = 2b = 20

ncert solution

Length of latus rectumncert solution

Question7. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 36x2 + 4y= 144

Solution :
The given equation is 36x2 + 4y2 = 144.

It can be written as

ncert solution

Here, the denominator of ncert solutionis greater than the denominator of ncert solution.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing equation (1) withncert solution, we obtain = 2 and a = 6.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (0, ±6).

Length of major axis = 2= 12

Length of minor axis = 2b = 4

ncert solution

Length of latus rectum ncert solution

Question8. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16x2 + y2 = 16

Solution :

The given equation is 16x2 + y2 = 16.

It can be written as

ncert solution

Here, the denominator of ncert solutionis greater than the denominator ofncert solution.

Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing equation (1) withncert solution, we obtain = 1 and a = 4.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (0, ±4).

Length of major axis = 2a = 8

Length of minor axis = 2b = 2

ncert solution

Length of latus rectumncert solution

Question 9. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 4x2 + 9y2 = 36

Solution :

The given equation is 4x2 + 9y2 = 36.

It can be written as

ncert solution

Here, the denominator of ncert solution is greater than the denominator of ncert solution.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with ncert solution, we obtain = 3 and b = 2.

ncert solution

Therefore,

The coordinates of the foci arencert solution.

The coordinates of the vertices are (±3, 0).

Length of major axis = 2a = 6

Length of minor axis = 2b = 4

ncert solution

Length of latus rectumncert solution

In each of the Exercises 10 to 20, find the equation of the ellipse that satisfies the given conditions:

Question 10. Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci (±4, 0)

Solution :
Vertices (±5, 0), foci (±4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where a is the semi-major axis.

Accordingly, a = 5 and c = 4.

It is known that chapter 11-Conic Sections Exercise 11.3.

chapter 11-Conic Sections Exercise 11.3

Thus, the equation of the ellipse ischapter 11-Conic Sections Exercise 11.3.

Question11. Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13), foci (0, ±5)

Solution :

Vertices (0, ±13), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the ellipse will be of the form  chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, a = 13 and c = 5.

It is known that chapter 11-Conic Sections Exercise 11.3.

ncert solution

Thus, the equation of the ellipse is ncert solution.

Question 12. Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci (±4, 0)

Solution :

Vertices (±6, 0), foci (±4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, a = 6, c = 4.

It is known that chapter 11-Conic Sections Exercise 11.3.

ncert solution

Thus, the equation of the ellipse is ncert solution

Question 13.Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Solution :

Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Here, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, a = 3 and b = 2.

Thus, the equation of the ellipse is ncert solution

Question 14. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ± √5), ends of minor axis (± 1, 0)

Solution :

Ends of major axis(0, ± √5), ends of minor axis (±1, 0)

Here, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, a =5 and b = 1.

Thus, the equation of the ellipse is ncert solution

Question 15. Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (± 5, 0)

Solution :

Length of major axis = 26; foci = (±5, 0).

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, 2a = 26 ⇒ a = 13 and c = 5.

It is known that chapter 11-Conic Sections Exercise 11.3.

ncert solution

Thus, the equation of the ellipse is ncert solution.

.

Question 16. Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ± 6).

Solution :

Length of minor axis = 16; foci = (0, ±6).

Since the foci are on the y-axis, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, 2b = 16 ⇒ b = 8 and c = 6.

It is known that chapter 11-Conic Sections Exercise 11.3.

ncert solution

Thus, the equation of the ellipse is ncert solution.

.

Question 17. Find the equation for the ellipse that satisfies the given conditions: Foci (±3, 0), a = 4

Solution :

Foci (±3, 0), a = 4

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, c = 3 and a = 4.

It is known that chapter 11-Conic Sections Exercise 11.3.

ncert solution

Thus, the equation of the ellipse is ncert solution

Question 18. Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the x axis.

Solution :

It is given that b = 3, c = 4, centre at the origin; foci on the axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form chapter 11-Conic Sections Exercise 11.3, where is the semi-major axis.

Accordingly, b = 3, c = 4.

It is known thatchapter 11-Conic Sections Exercise 11.3.

ncert solution

Thus, the equation of the ellipse is ncert solution

Question 19. Find the equation for the ellipse that satisfies the given conditions:  Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Solution :

Since the centre is at (0, 0) and the major axis is on the y-axis, the equation of the ellipse will be of the form

ncert solution

The ellipse passes through points (3, 2) and (1, 6). Hence,

ncert solution

On solving equations (2) and (3), we obtain b2 = 10 and a2 = 40.

Thus, the equation of the ellipse isncert solution

Question 20. Major axis on the x- axis and passes through the points (4, 3) and (6, 2).

Solution :

Since the major axis is on the x-axis, the equation of the ellipse will be of the form

ncert solution

The ellipse passes through points (4, 3) and (6, 2). Hence,

ncert solution

On solving equations (2) and (3), we obtain a2 = 52 and b2 = 13.

Thus, the equation of the ellipse isncert solution.