Answer:a) The equation is (y - 1)² = -8 (x - 4)b) The equation is (x - 1)²/25 + (y - 4)²/16 = 1c) The equation of the ellipse is (x - 3)²/16 + y²/4 = 1Step-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 = 1c) 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