A hyperbola whose centre is (2,3) has its foci √13 units away from the centre along an axis parallel to the y-axis. Given further that the vertices are 2 units away from the centre... A hyperbola whose centre is (2,3) has its foci √13 units away from the centre along an axis parallel to the y-axis. Given further that the vertices are 2 units away from the centre, find: (i) The equation of the hyperbola in the form Ax² + Bxy + Cy² + Dx + Ey + F = 0 (ii) State the equation of the asymptotes on the XY plane. A curve is defined by x² + y² - 8y - 6x = 0. (i) Show that the curve defines a circle. (ii) Locate the centre C and radius r of the circle. Read more

Steps to Solve

1. Identify Parameters of the Hyperbola

The center is given as ( (h, k) = (2, 3) ).
The distance from the center to the foci is ( c = \sqrt{13} ) and the distance to the vertices is ( a = 2 ).
Since the foci are along the y-axis, we note that ( b ) can be found using the relationship ( c^2 = a^2 + b^2 ):
$$ c^2 = a^2 + b^2 $$
$$ 13 = 4 + b^2 $$
Thus,
$$ b^2 = 9 $$
and
$$ b = 3 $$

2. Write the Standard Form of the Hyperbola

The standard form of a hyperbola centered at ( (h, k) ) with a vertical transverse axis is:
$$ \frac{(x - h)^2}{b^2} - \frac{(y - k)^2}{a^2} = -1 $$
Substituting ( h, k, a, b ):
$$ \frac{(x - 2)^2}{9} - \frac{(y - 3)^2}{4} = -1 $$

3. Transform to General Form

Multiply through by -1:
$$ -\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1 $$
Now, multiply both sides by 36 (LCM of 9 and 4):
$$ -4(x - 2)^2 + 9(y - 3)^2 = 36 $$
Expand:
$$ -4(x^2 - 4x + 4) + 9(y^2 - 6y + 9) = 36 $$
Distributing:
$$ -4x^2 + 16x - 16 + 9y^2 - 54y + 81 = 36 $$
Combine:
$$ -4x^2 + 9y^2 + 16x - 54y + 49 = 0 $$

4. Find the Asymptotes

The equations of the asymptotes for a vertical hyperbola are given by:
$$ y - k = \pm \frac{a}{b} (x - h) $$
Here, ( \frac{a}{b} = \frac{2}{3} ):
Thus, the equations are:
$$ y - 3 = \frac{2}{3}(x - 2) $$
$$ y - 3 = -\frac{2}{3}(x - 2) $$
Solving these equations gives the asymptotes:

1. ( y = \frac{2}{3}x + \frac{5}{3} )

2. ( y = -\frac{2}{3}x + \frac{13}{3} )

3. Show that the Curve Defines a Circle

Given curve is:
$$ x^2 + y^2 - 8y - 6x = 0 $$
Rearranging gives:
$$ x^2 - 6x + y^2 - 8y = 0 $$

6. Complete the Square

For ( x ):
$$ x^2 - 6x = (x - 3)^2 - 9 $$

For ( y ):
$$ y^2 - 8y = (y - 4)^2 - 16 $$

Substituting:
$$ (x - 3)^2 - 9 + (y - 4)^2 - 16 = 0 $$
$$ (x - 3)^2 + (y - 4)^2 = 25 $$

7. Identify Center and Radius of the Circle

The equation ( (x - 3)^2 + (y - 4)^2 = 5^2 ) defines a circle:

• Center ( C = (3, 4) )
• Radius ( r = 5 )