Answer:
a) The equation is (y - 1)² = -8 (x - 4)
b) The equation is (x - 1)²/25 + (y - 4)²/16 = 1
c) The equation of the ellipse is (x - 3)²/16 + y²/4 = 1
Step-by-step explanation:
a) Lets revise the standard form of the equation of the parabola with a
horizontal axis
# (y - k)² = 4p (x - h), (h , k) are the coordinates of its vertex and p ≠ 0
- The focus of it is (h + p , k)
* Lets solve the problem
∵ The focus is (2 , 1)
∵ focus is (h + p , k)
∴ h + p = 2 ⇒ subtract p from both sides
∴ h = 2 - p ⇒ (1)
∴ k = 1
∵ It opens left, then the axis is horizontal and p is negative
∴ Its equation is (y - k)² = 4p (x - h)
∵ k = 1
∴ Its equation is (y - 1)² = 4p (x - h)
- The parabola contains point (2 , 5), substitute the coordinates of the
point in the equation of the parabola
∴ (5 - 1)² = 4p (2 - h)
∴ (4)² = 4p (2 - h)
∴ 16 = 4p (2 - h) ⇒ divide both sides by 4
∴ 4 = p (2 - h) ⇒ (2)
- Use equation (1) to substitute h in equation (2)
∴ 4 = p (2 - [2 - p]) ⇒ open the inside bracket
∴ 4 = p (2 - 2 + p) ⇒ simplify
∴ 4 = p (p)
∴ 4 = p² ⇒ take √ for both sides
∴ p = ± 2, we will chose p = -2 because the parabola opens left
- Substitute the value of p in (1) to find h
∵ h = 2 - p
∵ p = -2
∴ h = 2 - (-2) = 2 + 2 = 4
∴ The equation of the parabola in standard form is
(y - 1)² = 4(-2) (x - 4)
∴ The equation is (y - 1)² = -8 (x - 4)
b) Lets revise the equation of the ellipse
- The standard form of the equation of an ellipse with center (h , k)
and major axis parallel to x-axis is (x - h)²/a² + (y - k)²/b² = 1
- The coordinates of the vertices are (h ± a , k )
- The coordinates of the foci are (h ± c , k), where c² = a² - b²
* Now lets solve the problem
∵ Its vertices are (-4 , 4) and (6 , 4)
∵ The coordinates of the vertices are (h + a , k ) and (h - a , k)
∴ k = 4
∴ h + a = 6 ⇒ (1)
∴ h - a = -4 ⇒ (2)
- Add (1) and (2) to find h
∴ 2h = 2 ⇒ divide both sides by 2
∴ h = 1
- Substitute the value of h in (1) or (2) to find a
∴ 1 + a = 6 ⇒subtract 1 from both sides
∴ a = 5
∵ The foci at (-2 , 4) and (4 , 4)
∵ The coordinates of the foci are (h + c , k) , (h - c , k)
∴ h + c = 4
∵ h = 1
∴ 1 + c = 4 ⇒ subtract 1 from both sides
∴ c = 3
∵ c² = a² - b²
∴ 3² = 5² - b²
∴ 9 = 25 - b² ⇒ subtract 25 from both sides
∴ -16 = -b² ⇒ multiply both sides by -1
∴ 16 = b²
∵ a² = 25
∵ The equation of the ellipse is (x - h)²/a² + (y - k)²/b² = 1
∴ The equation is (x - 1)²/25 + (y - 4)²/16 = 1
c) How to identify the type of the conic
- Rewrite the equation in the general form,
Ax² + Bxy + Cy² + Dx + Ey + F = 0
- Identify the values of A and C from the general form.
- If A and C are nonzero, have the same sign, and are not equal
to each other, then the graph is an ellipse.
- If A and C are equal and nonzero and have the same sign, then
the graph is a circle
- If A and C are nonzero and have opposite signs, and are not equal
then the graph is a hyperbola.
- If either A or C is zero, then the graph is a parabola
* Now lets solve the problem
∵ x² + 4y² - 6x - 7 = 0
∵ The general form of the conic equation is
Ax² + Bxy + Cy² + Dx + Ey + F = 0
∴ A = 1 and C = 4
∵ If A and C are nonzero, have the same sign, and are not equal to
each other, then the graph is an ellipse.
∵ x² + 4y² - 6x - 7 = 0 ⇒ re-arrange the terms
∴ (x² - 6x ) + 4y² - 7 = 0
- Lets make x² - 6x completing square
∵ 6x ÷ 2 = 3x
∵ 3x = x × 3
- Lets add and subtract 9 to x² - 6x to make the completing square
x² - 6x + 9 = (x - 3)²
∴ (x² - 6x + 9) - 9 + 4y² - 7 = 0 ⇒ simplify
∴ (x - 3)² + 4y² - 16 = 0 ⇒ add 16 to both sides
∴ (x - 3)² + 4y² = 16 ⇒ divide all terms by 16
∴ (x - 3)²/16 + 4y²/16 = 1 ⇒ simplify
∴ (x - 3)²/16 + y²/4 = 1
∴ The equation of the ellipse is (x - 3)²/16 + y²/4 = 1