Answer to Question #219588 in Analytic Geometry for Bless

Question

Express the following equations of a parabola in standard form and in each case state the coordinates of its vertex,focus and the ends of the latus rectum.

(a) x^2− 2x − 4y = 0,

(b) y^2 + 12x − 48 = 0,

(c) x^2 − 6x − 2y + 7 = 0.

Answer 1

(a)

$$x^2-2x+1=4y+1$$

$$(x-1)^2=4\cdot1(y+\dfrac{1}{4})$$

$p=1>0$

$h=1, k=-\dfrac{1}{4}$

$$k+p=-\dfrac{1}{4}+1=\dfrac{3}{4}$$

Vertex: $V(1, -\dfrac{1}{4})$

Focus: $F(1, \dfrac{3}{4})$

Latus rectum: $y=\dfrac{3}{4}$

$$x^2-2x+1=4(\dfrac{3}{4})+1$$

$$x^2-2x-3=0$$

$$(x+1)(x-3)=0$$

$$x_1=-1, x_2=3$$

The ends of the latus rectum

$$(-1, \dfrac{3}{4}), (3, \dfrac{3}{4})$$

(b)

$$y^2 + 12x − 48 = 0$$

$$y^2=-4\cdot3(x-4)$$

$p=-3<0$

$h=4, k=0$

$$h+p=4-3=1$$

Vertex: $V(4, 0)$

Focus: $F(1, 0)$

Latus rectum: $x=1$

$$y^2 + 12(1) − 48 = 0$$

$$y^2=36$$

$$y_1=-6, y_2=6$$

The ends of the latus rectum

$$(1, -6), (1, 6)$$

(c)

$$x^2 − 6x − 2y + 7 = 0$$

$$x^2-6x+9=2y+2$$

$$(x-3)^2=4\cdot(\dfrac{1}{2})(y+1)$$

$p=\dfrac{1}{2}>0$

$h=3, k=-1$

$$k+p=-1+\dfrac{1}{2}=-\dfrac{1}{2}$$

Vertex: $V(3, -1)$

Focus: $F(3, -\dfrac{1}{2})$

Latus rectum: $y=-\dfrac{1}{2}$

$$x^2 − 6x − 2(-\dfrac{1}{2}) + 7 =0$$

$$x^2-6x+8=0$$

$$(x-2)(x-4)=0$$

$$x_1=2, x_2=4$$

The ends of the latus rectum

$$(2, -\dfrac{1}{2}), (3, -\dfrac{1}{2})$$