Parabola – Locus II By Mr Porter

Assumed Knowledge: Simple Examples Students need to know, understand and apply the skills to complete the square of a quadratic. Simple Examples a) x2 + 8x + 12 = 0 Rearrange to x2 + bx = c b) Divide by 2 and rearrange to x2 + bx = c x2 + 8x = -12 For ax2 + 2bx + b2 = 0 2b = (+8) For ax2 + 2bx + b2 = 0 2b = (+4) To complete the square we need to add b2 = +16 to both sides. To complete the square we need to add b2 = +4 to both sides. x2 + 8x + 16 = -12 + 16 x2 + 8x + 16 = 4 Factorise LHS. (x + 4)2 = 4 Take √ of both sides. Take √ of both sides. x + 4 = ±2 Solve for x. Solve for x. x = -4±2 i.e. x = -4 - 2 or x = -4 + 2 x = -6 or x = -2 The solution is x = -6 and x = -2.

Example 1: Rewrite y = x2 – 4x + 6 in the form (x – h)2 = 4a(y – k) and hence state the vertex, focus and directrix of the parabola. Solution: So, Vertex (h, k) = (2, 2), focal length a = ¼. Rearrange with all x-terms on the right and everything else on the left side of the ‘=‘. Focus (h, k +a) = (2, 2 + ¼) x2 – 4x + 6 = y Parabola is concave-up! x2 – 4x = y – 6 Now, “complete the square” for ‘x’. Directrix y = k – a x2 – 4x + 4 = y – 6 + 4 Factorise RHS. y = 2 – ¼ (x – 2)2= y – 2 Write in the form (x – h)2 = 4a(y – k) (x – 2)2= 1(y – 2) (x – h)2 = 4a(y – k), vertex at (h, k)

Example 2: Rewrite y = 2x2 + 8x – 10 in the form (x – h)2 = 4a(y – k) and hence state the vertex, focus and directrix of the parabola. Solution: So, Vertex (h, k) = (-2, -18), focal length a = 1/8, concave up (+). Rearrange with all x-terms on the left and everything else on the right side of the ‘=‘. Focus (h, k +a) = (-2, -18 + 1/8) 2x2 + 8x – 10 = y Parabola is concave-up! 2x2 + 8x = y + 10 Now, “complete the square” for ‘x’. Directrix y = k – a x2 + 4x = 1/2 y +5 y = -18 – 1/8 x2 + 4x + 4 = 1/2 y +5 + 4 Factorise LHS. (x + 2)2= 1/2 y + 9 Write in the form (x – h)2 = 4a(y – k) (x + 2)2= 1/2 (y + 18) (x – -2)2= 1/2 (y – -18) (x – h)2 = 4a(y – k), Vertex at (h, k)

Example 3: For the parabola, y = –x2 + 6x – 10 , state the vertex, focus and write the equation of the directrix. Solution: So, Vertex (h, k) = (3, -1), focal length a = 1/4,, concave down (-). Rearrange with all x-terms on the left and everything else on the right side of the ‘=‘. Focus (h, k +a) = (3, -1 – 1/4) –x2 + 6x – 10 = y Parabola is concave-down! x2 – 6x = –y – 10 Rearrange and “complete the square” for ‘x’. Directrix y = k + a x2 – 6x + 9 = –y – 10 + 9 y = -1 + 1/4 x2 – 6x + 9 = –y – 1 Factorise LHS. (x – 3)2 = –y – 1 Write in the form (x – h)2 = –4a(y – k) (x – 3)2 = –(y + 1) (x – 3)2 = –(y – -1) (x – h)2 = -4a(y – k), Vertex at (h, k)

Example 4: For the parabola, y = – 1/4 x2 + 2x – 2 , state the vertex, focus and write the equation of the directrix and the length of the latus rectum. Solution: So, Vertex (h, k) = (4, 1), focal length a = 1, concave down (-). Rearrange with all x-terms on the left and everything else on the right side of the ‘=‘. Parabola is concave-down!Multiply everything by ‘4’. Focus (h, k +a) = (4, 1 – 1) –¼ x2 + 2x – 2 = y = (4, 0) x2 – 8x = –4y – 8 Rearrange and “complete the square” for ‘x’. Directrix y = k + a x2 – 8x + 16 = –4y – 8 + 16 y = 1 + 1 x2 – 8x + 16 = –4y + 8 Factorise LHS. y = 2 (x – 4)2 = –4y + 8 Write in the form (x – h)2 = –4a(y – k) Length of latus rectum = |4a| (x – 4)2 = –4(y – 1) = 4 (x – h)2 = -4a(y – k), Vertex at (h, k)

Example 5: For the parabola, x = y2 + 2y – 7 , state the vertex, focus and write the equation of the directrix and the length of the latus rectum. Solution: So, Vertex (h, k) = (-1, -8), focal length a = ¼, concave right (+) . Rearrange with all y-terms on the right and everything else on the left side of the ‘=‘. Focus (h +a, k) = (-1 + ¼ , -8) y2 + 2y – 7 = x Parabola is concave-right! = (-3/4 , -8) y2 + 2y = x + 7 Rearrange and “complete the square” for ‘y’. Directrix x = h – a y2 + 2y + 1 = x + 7 + 1 x = -1 – ¼ y2 + 2y + 1 = x + 8 Factorise LHS. x = - 5/4 (y +1)2 = x + 8 Write in the form (y – k)2 = 4a(x – h) 4x + 5 = 0 (y + 1)2 = 1(x + 8) (y – -1)2 = 1(x – -8) Length of latus rectum = |4a| = 1 (y – k)2 = 4a(x – h), vertex at (h, k)

Example 6: For the parabola, x = – 1/3 y2 – 2y + 9 , state the vertex, focus and write the equation of the directrix and the length of the latus rectum. Solution: So, Vertex (h, k) = (-3, -12), Focal length ‘a’ = 3/4 , concave left (-) Rearrange with all y-terms on the right and everything else on the left side of the ‘=‘. Parabola is concave-left! Multiply everthing by ‘-3’. Focus (h + a, k) = (-3 + – 3/4 , -12) –1/3 y2 – 2y + 9 = x = (-33/4 , -12) y2 + 6y = -3x + 27 Rearrange and “complete the square” for ‘y’. Directrix x = h – a y2 + 6y + 9 = -3x + 27 + 9 x = -3 – - 3/4 y2 + 6y + 9 = -3x + 36 Factorise LHS. x = -15/4 (y + 3)2 = -3(x + 12) Write in the form (y – k)2 = -4a(x – h) 4x + 15 = 0 (y + 3)2 = -3(x + 12) (y + -3)2 = -4( 3/4 )(x – -12) Length of latus rectum = |4a| = 3 (y – k)2 = 4a(x – h), vertex at (h, k)