The extremities of latus rectum of a parabola are (1,1) and (1,-1). Then the equation of the parabola can be

Question

The extremities of latus rectum of a parabola are (1,1) and (1,-1). Then the equation of the parabola can be (A) y2=2 x-1 (B) y2=1-2 x (C) y2=3-2 x (D) y2=2 x-4

Tardigrade Admin · Accepted Answer

Given that the extremities of the latus rectum are

(1, 1)

and

(1, - 1)

. Then,

4 a = 2

or

a = \frac{1}{2}

Also, the focus of the parabola is

(1, 0)

.
Hence, the vertex can be

(1/2, 0)

or

(3/2, 0)

.
Therefore, the equations of the parabola can be

y^{2} = 2 (x - \frac{1}{2})

or

y^{2} = - 2 (x - \frac{3}{2})

i.e.,

y^{2} = 2 x - 1

or

y^{2} = 3 - 2 x

Q. The extremities of latus rectum of a parabola are $(1, 1)$ and $(1, - 1)$ . Then the equation of the parabola can be

$y^{2} = 2 x - 1$

$y^{2} = 1 - 2 x$

$y^{2} = 3 - 2 x$

$y^{2} = 2 x - 4$

Q. The extremities of latus rectum of a parabola are (1,1) and (1,−1). Then the equation of the parabola can be

y2=2x−1

y2=1−2x

y2=3−2x

y2=2x−4

Given that the extremities of the latus rectum are (1,1) and (1,−1). Then, 4a=2 or a=21​ Also, the focus of the parabola is (1,0). Hence, the vertex can be (1/2,0) or (3/2,0). Therefore, the equations of the parabola can be y2=2(x−21​) or y2=−2(x−23​) i.e., y2=2x−1 or y2=3−2x

Q. The extremities of latus rectum of a parabola are $(1, 1)$ and $(1, - 1)$ . Then the equation of the parabola can be

$y^{2} = 2 x - 1$

$y^{2} = 1 - 2 x$

$y^{2} = 3 - 2 x$

$y^{2} = 2 x - 4$