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

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