How do you graph x^2 - y^2 - 4x + 6y - 3 = 0.?

Question

his is a circle graph, I'm having some trouble. I need the center point on a cartesian plane and the radius

Answer 1

Identify conic section :

If the given equation is in the form of Ax² + Bxy + Cy² + Dx + Ey + F = 0, where B = 0,represented by looking at A and C.

Conic section	Relationship of A and C
Parabola	A = 0 or C = 0 but not both
Circle	A = C
Ellipse	A and C have the same sign and A ≠ C
Hyperbola	A and C have opposite signs.

Standard form of equation	(x - h)2/a2 - (y - k)2/b2 = 1	(y - k)2/a2 - (x - h)2/b2 = 1
Direction of transverse axis	horizontal	vertical
Equations of asymptotes	y = k ± [(b/a)(x - h)]	y = k ± [(a/b)(x - h)]

If the hyperbola is horizontal :  .

• Where, "a " is the number in the denominator of the positive term, If the x - term is positive, then the hyperbola is horizontal
• a = semi - transverse axis , b = semi - conjugate axis
• Center: (h, k )
• Vertices: (h + a, k ), (h - a, k )
• Foci: (h + c, k ), (h - c, k )

If the hyperbola is vertical : (y - k)²/a² - (x - h)²/b² = 1.

• Where, "b " is the number in the denominator of the positive term, If the x - term is negative, then the hyperbola is vertical.
• a = semi - transverse axis , b = semi - conjugate axis .
• Center: (h, k )
• Vertices: (h , k + a ) and (h, k - a).
• Foci: (h , k +c) and (h, k - c).

The equation is x² - y² - 4x + 6y - 3 = 0.

Compare the above equation with Ax² + Bxy + Cy² + Dx + Ey + F = 0.

A = 1 and C = - 1.

Since, A and C have opposite signs The given equation represents the curve hyperbola.

Write the equation : x² - y² - 4x + 6y - 3 = 0 in standard form of hyperbola : .

x² - y² - 4x + 6y - 3 = 0

x² - 4x - 3 = y² - 6y.

To change the expression into a perfect square  add (half the x coefficient)² and (half the y - coeffient)² to each side of the expression.

Here x coefficient = - 4, so, (half the x coefficient)² = (- 4/2)²= 4.

Here y coefficient = - 6, so, (half the x coefficient)² = (- 6/2)²= 9.

Add 4 and 9 to each side of the equation.

x² - 4x + 4 + 9 - 3 = y² - 6y + 9 + 4

(x - 2)² + 6 = (y - 3)² + 4

(x - 2)² - (y - 3)² = 4 - 6 = - 2

- (x - 2)²/2 + (y - 3)²/2 = 1

(y - 3)²/2 - (x - 2)²/2 = 1

(y - 3)²/(√2)² - (x - 2)²/(√2)² = 1.

Answer 2

Contd.......

Compare the above equation with standard form of vertical hyperbola : (y - k)²/a² - (x - h)²/b² = 1.

a = semi - transverse axis = √2,

b = semi - conjugate axis = √2,

Center: (h, k ) = (2, 3),

Vertices: (h , k + a ) and (h, k - a) = (2, 3 + √2) and (2, 3 + √2).

c² = a²+ b².

c² = √2² + √2²

= 2 + 2 = 4

⇒ c = √4 = 2.

Foci: (h , k +c) and (h, k - c) : (2, 5) and (2, 1).

 Asympototes of hyperbola are : y = k ± [(a/b)(x - h)].

y = 3 ± [(√4/√4)(x - 2)]

y = 3 ± (x - 2).

Graph of hyperbola:

• Draw the coordinate plane.
• Plot the center of hyperbola (2,3).
• To graph the hyperbola go √4 units up and down from center point and √4 units left and right from center point.
• Use these points to draw a rectangle .
• Draw diagonal lines through the center and the corner of the rectangle. These are asymptotes.
• The graph approaches the asymptotes but never actually touches them.
• Draw the curves, beginning at each vertex separately, that hug the asymptotes the farther away from the vertices the curve gets.
• Plot the vertices and foci of hyperbola.