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Exercise 11.4 - Chapter 11 Conic Sections class 11 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbolahttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4830/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_42610b97.gif

Answer:

The given equation ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4830/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_414f24f2.gif .

On comparing this equation with the standard equation of hyperbola i.e.,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4830/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_452a20d9.gif , we obtain a = 4 and b = 3.

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4830/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_3a003c40.gif

Therefore,

The coordinates of the foci are (±5, 0).

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4830/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_17815d06.gif

Length of latus rectumhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4830/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_3818a1a6.gif

Question 2:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbolahttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4832/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_31f97c09.gif

Answer:

The given equation ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4832/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_11ebf286.gif .

On comparing this equation with the standard equation of hyperbola i.e.,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4832/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif , we obtain a = 3 andhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4832/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m73892847.gif .

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4832/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_7f2f98f7.gif

Therefore,

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

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4832/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m7c9be18d.gif

Length of latus rectumhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4832/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m37d73f3d.gif

Question 3:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36

Answer:

The given equation is 9y2 – 4x2 = 36.

It can be written as

9y2 – 4x2 = 36

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4835/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m718e4008.gif

On comparing equation (1) with the standard equation of hyperbola i.e.,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4835/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif , we obtain a = 2 and b = 3.

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4835/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1f211a6e.gif

Therefore,

The coordinates of the foci arehttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4835/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_11e4896a.gif .

The coordinates of the vertices arehttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4835/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_7efdea5e.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4835/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m55cc08b2.gif

Length of latus rectumhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4835/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_42d4563b.gif

Question 4:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x2 – 9y2 = 576

Answer:

The given equation is 16x2 – 9y2 = 576.

It can be written as

16x2 – 9y2 = 576

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4836/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m509d93f4.gif

On comparing equation (1) with the standard equation of hyperbola i.e.,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4836/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_452a20d9.gif , we obtain a = 6 and b = 8.

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4836/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_28c9ab25.gif

Therefore,

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

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4836/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_55ec5792.gif

Length of latus rectumhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4836/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_75193562.gif

Question 5:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 – 9x2 = 36

Answer:

The given equation is 5y2 – 9x2 = 36.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m17bd206f.gif

On comparing equation (1) with the standard equation of hyperbola i.e.,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif , we obtain a = https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_md38009f.gif  and b = 2.

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1a4da293.gif

Therefore, the coordinates of the foci arehttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_37cecc83.gif .

The coordinates of the vertices arehttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_7b4d72ee.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m209a3f40.gif

Length of latus rectumhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4837/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1bc2151f.gif

Question 6:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y2 – 16x2 = 784

Answer:

The given equation is 49y2 – 16x2 = 784.

It can be written as 49y2 – 16x2 = 784

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4838/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m2b8a10f.gif

On comparing equation (1) with the standard equation of hyperbola i.e.,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4838/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif , we obtain a = 4 and b = 7.

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4838/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_53a5dc98.gif

Therefore,

The coordinates of the foci arehttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4838/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_240316c5.gif .

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4838/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_135f67fc.gif

Length of latus rectumhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4838/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_ma313c9b.gif

Question 7:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)

Answer:

Vertices (±2, 0), foci (±3, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4839/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_452a20d9.gif .

Since the vertices are (±2, 0), = 2.

Since the foci are (±3, 0), c = 3.

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4839/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_3e4f7397.gif

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4839/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1419729.gif .

Question 8:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8)

Answer:

Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4840/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif .

Since the vertices are (0, ±5), = 5.

Since the foci are (0, ±8), c = 8.

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4840/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_14432639.gif

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4840/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_710bfc03.gif .

Question 9:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5)

Answer:

Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4842/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif .

Since the vertices are (0, ±3), = 3.

Since the foci are (0, ±5), c = 5.

We know that a2 + b2 = c2.

∴32 + b2 = 52

⇒ b2 = 25 – 9 = 16

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4842/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_58b45b65.gif .

Question 10:

Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.

Answer:

Foci (±5, 0), the transverse axis is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4843/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_452a20d9.gif .

Since the foci are (±5, 0), c = 5.

Since the length of the transverse axis is 8, 2a = 8 ⇒ a = 4.

We know that a2 + b2 = c2.

∴42 + b2 = 52

⇒ b2 = 25 – 16 = 9

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4843/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m511f9386.gif .

Question 11:

Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.

Answer:

Foci (0, ±13), the conjugate axis is of length 24.

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4844/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif .

Since the foci are (0, ±13), c = 13.

Since the length of the conjugate axis is 24, 2b = 24 ⇒ b = 12.

We know that a2 + b2 = c2.

a2 + 122 = 132

⇒ a2 = 169 – 144 = 25

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4844/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1b9c159e.gif .

Question 12:

Find the equation of the hyperbola satisfying the give conditions: Focihttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4845/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1b661338.gif , the latus rectum is of length 8.

Answer:

Focihttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4845/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1b661338.gif , the latus rectum is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4845/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_452a20d9.gif .

Since the foci arehttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4845/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_1b661338.gifc =https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4845/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_6feae8dd.gif .

Length of latus rectum = 8

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4845/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_180fe2ef.gif

We know that a2 + b2 = c2.

a2 + 4a = 45

⇒ a2 + 4a – 45 = 0

⇒ a2 + 9a – 5a – 45 = 0

⇒ (a + 9) (a – 5) = 0

⇒ a = –9, 5

Since a is non-negative, = 5.

b2 = 4= 4 × 5 = 20

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4845/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_18e41b56.gif .

Question 13:

Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12

Answer:

Foci (±4, 0), the latus rectum is of length 12.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4846/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_452a20d9.gif .

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4846/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m76c7b892.gif

We know that a2 + b2 = c2.

a2 + 6a = 16

⇒ a2 + 6a – 16 = 0

⇒ a2 + 8a – 2a – 16 = 0

⇒ (a + 8) (a – 2) = 0

⇒ a = –8, 2

Since a is non-negative, = 2.

b2 = 6= 6 × 2 = 12

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4846/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m3604f7d3.gif .

Question 14:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4847/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_4d4605ae.gif

Answer:

Vertices (±7, 0), https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4847/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_4d4605ae.gif

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4847/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_452a20d9.gif .

Since the vertices are (±7, 0), = 7.

It is given that https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4847/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_4d4605ae.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4847/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_741f8acc.gif

We know that a2 + b2 = c2.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4847/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_21dee499.gif

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4847/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_36974c71.gif .

Question 15:

Find the equation of the hyperbola satisfying the give conditions: Focihttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m7f61218.gif , passing through (2, 3)

Answer:

Focihttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m7f61218.gif , passing through (2, 3)

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m515f458b.gif .

Since the foci arehttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m7f61218.gifc =https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m1829d81a.gif .

We know that a2 + b2 = c2.

∴ a2 + b2 = 10

⇒ b2 = 10 – a2 … (1)

Since the hyperbola passes through point (2, 3),

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_14845639.gif

From equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_580a5c86.gif

In hyperbola, c > a, i.e., c2 > a2

∴ a2 = 5

⇒ b2 = 10 – a2 = 10 – 5 = 5

Thus, the equation of the hyperbola ishttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4848/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m7f01a2ea.gif .

Also Read : Miscellaneous-Exercise-Chapter-11-Conic-Sections-class-11-ncert-solutions-Maths

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