Chapter 10: Conic Sections

x²/16 − y²/9 = 1

Solutions:

The given equation has a positive x² term, so the transverse axis is along the x-axis.

Comparing with the standard equation x²/a² − y²/b² = 1, we get a² = 16 and b² = 9, so a = 4 and b = 3.

Now, c = √(a² + b²) = √(16 + 9) = √25 = 5.

Coordinates of the foci = (± 5, 0)

Coordinates of the vertices = (± 4, 0)

Eccentricity, e = c/a = 5/4

Length of the latus rectum = 2b²/a = (2 × 9)/4 = 9/2

y²/9 − x²/27 = 1

Solutions:

Here the positive term is y², so the transverse axis is along the y-axis.

Comparing with y²/a² − x²/b² = 1, we get a² = 9 and b² = 27, so a = 3 and b = 3√3.

Now, c = √(a² + b²) = √(9 + 27) = √36 = 6.

Coordinates of the foci = (0, ± 6)

Coordinates of the vertices = (0, ± 3)

Eccentricity, e = c/a = 6/3 = 2

Length of the latus rectum = 2b²/a = (2 × 27)/3 = 18

9y² − 4x² = 36

Solutions:

Dividing both sides by 36: y²/4 − x²/9 = 1.

The transverse axis is along the y-axis. Here a² = 4, b² = 9, so a = 2 and b = 3.

Now, c = √(a² + b²) = √(4 + 9) = √13.

Coordinates of the foci = (0, ± √13)

Coordinates of the vertices = (0, ± 2)

Eccentricity, e = c/a = √13/2

Length of the latus rectum = 2b²/a = (2 × 9)/2 = 9

16x² − 9y² = 576

Solutions:

Dividing both sides by 576: x²/36 − y²/64 = 1.

The transverse axis is along the x-axis. Here a² = 36, b² = 64, so a = 6 and b = 8.

Now, c = √(a² + b²) = √(36 + 64) = √100 = 10.

Coordinates of the foci = (± 10, 0)

Coordinates of the vertices = (± 6, 0)

Eccentricity, e = c/a = 10/6 = 5/3

Length of the latus rectum = 2b²/a = (2 × 64)/6 = 64/3

5y² − 9x² = 36

Solutions:

5. 5y² − 9x² = 36

Dividing both sides by 36: y²/(36/5) − x²/4 = 1.

The transverse axis is along the y-axis. Here a² = 36/5 and b² = 4, so a = 6/√5.

Now, c² = a² + b² = 36/5 + 4 = 56/5, so c = √(56/5) = 2√70/5.

Coordinates of the foci = (0, ± 2√70/5)

Coordinates of the vertices = (0, ± 6/√5), i.e., (0, ± 6√5/5)

Eccentricity, e = c/a = (2√70/5) ÷ (6/√5) = √14/3

Length of the latus rectum = 2b²/a = 8 ÷ (6/√5) = 8√5/6 = 4√5/3

49y² − 16x² = 784

Solutions:

Dividing both sides by 784: y²/16 − x²/49 = 1.

The transverse axis is along the y-axis. Here a² = 16, b² = 49, so a = 4 and b = 7.

Now, c = √(a² + b²) = √(16 + 49) = √65.

Coordinates of the foci = (0, ± √65)

Coordinates of the vertices = (0, ± 4)

Eccentricity, e = c/a = √65/4

Length of the latus rectum = 2b²/a = (2 × 49)/4 = 49/2

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

Vertices (± 2, 0), foci (± 3, 0)

Solutions:

Since the vertices and foci lie on the x-axis, the equation is of the form x²/a² − y²/b² = 1.

From the vertices, a = 2; from the foci, c = 3.

Then b² = c² − a² = 9 − 4 = 5.

Required equation: x²/4 − y²/5 = 1

Vertices (0, ± 5), foci (0, ± 8)

Solutions:

Since the vertices and foci lie on the y-axis, the equation is of the form y²/a² − x²/b² = 1.

Here a = 5 and c = 8, so b² = c² − a² = 64 − 25 = 39.

Required equation: y²/25 − x²/39 = 1

Vertices (0, ± 3), foci (0, ± 5)

Solutions:

The foci lie on the y-axis, so the equation is of the form y²/a² − x²/b² = 1.

Here a = 3 and c = 5, so b² = c² − a² = 25 − 9 = 16.

Required equation: y²/9 − x²/16 = 1

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

Solutions:

The foci lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1.

Length of the transverse axis = 2a = 8, so a = 4. Also c = 5.

Then b² = c² − a² = 25 − 16 = 9.

Required equation: x²/16 − y²/9 = 1