Find the equation of the parabola whose focus is S(3, 5) and vertex is

1.	Find the equation of the parabola whose focus is S(3, 5) and vertex is A(1, 3).
Answer» Equation of the axis y – 3 = (3 - 5/ 1 - 3) (x - 1) = x – 1 x – y + 2 = 0 The directrix is perpendicular to the axis. Equation of the directrix is x + y + k = 0 Co-ordinates of Z be (x, y) A is the midpoint of SZ Co-ordinates of A are (3 + x/2, 5 + y/ 2) = (1, 3) 3 + x/2 = 1, 5 + y/2 = 3 3 + x = 2, 5 + y = 6 x = 2 – 3 = – 1, y = 6 – 5 = 1 Co-ordinates of Z are (–1, 1) The directrix passes through Z (–1, 1) – 1 + 1 + k = 0 ⇒ k = 0 Equation of the directrix is x – y = 0 Equation of the parabola is (x – α)² +(y – β)² (lx + my + n)²/ l² + m² (x – 3)² + (y – 5)² = (x + y)²/ 1 + 1 ⇒ 2(x² – 6x + 9 + y² – 10y + 25) = (x + y)² ⇒ 2x² + 2y² – 12x – 20y + 68 = x² + 2xy + y² i.e., x² – 2xy + y² – 12x – 20y + 68 = 0.

Find the equation of the parabola whose focus is S(3, 5) and vertex is A(1, 3).

Answer»

Equation of the axis y – 3 = (3 - 5/ 1 - 3) (x - 1)

= x – 1

x – y + 2 = 0

The directrix is perpendicular to the axis.

Equation of the directrix is x + y + k = 0

Co-ordinates of Z be (x, y)

A is the midpoint of SZ

Co-ordinates of A are

(3 + x/2, 5 + y/ 2) = (1, 3)

3 + x/2 = 1, 5 + y/2 = 3

3 + x = 2, 5 + y = 6

x = 2 – 3 = – 1, y = 6 – 5 = 1

Co-ordinates of Z are (–1, 1)

The directrix passes through Z (–1, 1)

– 1 + 1 + k = 0 ⇒ k = 0

Equation of the directrix is x – y = 0

Equation of the parabola is (x – α)² +(y – β)²

(lx + my + n)²/ l² + m²

(x – 3)² + (y – 5)² = (x + y)²/ 1 + 1

⇒ 2(x² – 6x + 9 + y² – 10y + 25) = (x + y)²

⇒ 2x² + 2y² – 12x – 20y + 68 = x² + 2xy + y²

i.e., x² – 2xy + y² – 12x – 20y + 68 = 0.

Discussion

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