What are the coordinates of the endpoints of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation is $(y - 1)^2 = 4(x + 1)$.Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get the vertex $(h, k) = (-1, 1)$ an

Vedclass · Accepted Answer

$(0, 3)$ and $(0, -1)$

Vedclass · Answer

(A) The given equation is $(y - 1)^2 = 4(x + 1)$.Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we get the vertex $(h, k) = (-1, 1)$ and $4a = 4$,so $a = 1$.The latus rectum of a parabola $Y^2 = 4aX$ has endpoints at $(a, 2a)$ and $(a, -2a)$.Here,$X = x - h = x + 1$ and $Y = y - k = y - 1$.Substituting $a = 1$,the endpoints in terms of $(X, Y)$ are $(1, 2)$ and $(1, -2)$.Now,converting back to $(x, y)$ coordinates:For $(1, 2)$: $x + 1 = 1 \implies x = 0$ and $y - 1 = 2 \implies y = 3$.For $(1, -2)$: $x + 1 = 1 \implies x = 0$ and $y - 1 = -2 \implies y = -1$.Thus,the endpoints are $(0, 3)$ and $(0, -1)$.

What are the coordinates of the endpoints of the latus rectum of the parabola $(y - 1)^2 = 4(x + 1)$?

Similar Questions

If a point $P$ moves such that its distances from the point $A(1, 1)$ and the line $x+y+2=0$ are equal,then the locus of $P$ is

Let $l$ be the directrix of the parabola $9y^2+12y+9x-14=0$ and $l_1$ be the line passing through the vertex of this parabola and the origin. If $(h, k)$ is the point of intersection of $l$ and $l_1$,then $h+k=$

If $\overrightarrow{AB}$ is the focal chord of the parabola $y^2=16x$ and $A=(1,-4)$,then the equation of the normal to the parabola at the point $B$ is

$P$ and $Q$ are two points on the parabola $y^2 = 8x$ and $S$ is its focus. $PS$ and $QS$ meet the curve again in $T$ and $R$ respectively. If $PQ$ passes through a fixed point $(-2, 3)$,then $TR$ also passes through a fixed point whose coordinates are

The focus of the parabola $y^2 = 4y - 4x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module