Chapter 10: Conic Sections

Ans: Since the vertices are on the y-axis, the major axis is along the y-axis, and the equation is of the form x²/b² + y²/a² = 1.

From the vertices (0, ±13), we get a = 13. From the foci (0, ±5), we get c = 5.

Using c² = a² − b²: 25 = 169 − b² b² = 169 − 25 = 144

Therefore, the equation of the ellipse is x²/144 + y²/169 = 1

Ans: Since the vertices are on the x-axis, the equation is of the form x²/a² + y²/b² = 1.

Here, a = 6 and c = 4.

Using c² = a² − b²: 16 = 36 − b² b² = 36 − 16 = 20

Therefore, the equation of the ellipse is x²/36 + y²/20 = 1

Ans: Since the ends of the major axis lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

From the ends of the major axis (±3, 0), we get a = 3. From the ends of the minor axis (0, ±2), we get b = 2.

Therefore, the equation of the ellipse is x²/9 + y²/4 = 1

Ans: Since the ends of the major axis lie on the y-axis, the major axis is along the y-axis, and the equation is of the form x²/b² + y²/a² = 1.

From the ends of the major axis (0, ±√5), we get a = √5, so a² = 5. From the ends of the minor axis (±1, 0), we get b = 1, so b² = 1.

Therefore, the equation of the ellipse is x²/1 + y²/5 = 1

Ans: Since the foci lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

Given, length of major axis = 2a = 26, so a = 13. From the foci (±5, 0), we get c = 5.

Using c² = a² − b²: 25 = 169 − b² b² = 144

Therefore, the equation of the ellipse is x²/169 + y²/144 = 1

Ans: Since the foci lie on the y-axis, the major axis is along the y-axis, and the equation is of the form x²/b² + y²/a² = 1.

Given, length of minor axis = 2b = 16, so b = 8. From the foci (0, ±6), we get c = 6.

Using c² = a² − b²: 36 = a² − 64 a² = 36 + 64 = 100

Therefore, the equation of the ellipse is x²/64 + y²/100 = 1

Ans: Since the foci lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

Given, c = 3 and a = 4.

Using c² = a² − b²: 9 = 16 − b² b² = 16 − 9 = 7

Therefore, the equation of the ellipse is x²/16 + y²/7 = 1

Ans: Since the foci lie on the x-axis, the major axis is along the x-axis, and the equation is of the form x²/a² + y²/b² = 1.

Given, b = 3 and c = 4.

Using the relation a² = b² + c²: a² = 9 + 16 = 25

Therefore, the equation of the ellipse is x²/25 + y²/9 = 1

Ans: Since the major axis is along the y-axis, the equation of the ellipse is of the form:

x²/b² + y²/a² = 1 ... (i)

Since the ellipse passes through the point (3, 2), substituting in equation (i): 9/b² + 4/a² = 1 ... (1)

Since the ellipse passes through the point (1, 6), substituting in equation (i): 1/b² + 36/a² = 1 ... (2)

Let 1/b² = p and 1/a² = q. Then the equations become: 9p + 4q = 1 ... (3) p + 36q = 1 ... (4)

From equation (4): p = 1 − 36q

Substituting in equation (3): 9(1 − 36q) + 4q = 1 9 − 324q + 4q = 1 −320q = −8 q = 1/40, which gives a² = 40

Now, p = 1 − 36 × (1/40) = 1 − 36/40 = 4/40 = 1/10, which gives b² = 10

Therefore, the equation of the ellipse is x²/10 + y²/40 = 1

Ans: Since the major axis is along the x-axis, the equation of the ellipse is of the form:

x²/a² + y²/b² = 1 ... (i)

Since the ellipse passes through the point (4, 3): 16/a² + 9/b² = 1 ... (1)

Since the ellipse passes through the point (6, 2): 36/a² + 4/b² = 1 ... (2)

Let 1/a² = p and 1/b² = q. Then: 16p + 9q = 1 ... (3) 36p + 4q = 1 ... (4)

Multiplying equation (3) by 4 and equation (4) by 9: 64p + 36q = 4 ... (5) 324p + 36q = 9 ... (6)

Subtracting equation (5) from equation (6): 260p = 5 p = 1/52, which gives a² = 52

Substituting p = 1/52 in equation (3): 16/52 + 9q = 1 9q = 1 − 4/13 = 9/13 q = 1/13, which gives b² = 13

Therefore, the equation of the ellipse is x²/52 + y²/13 = 1