The focus of the parabola y^2 - x - 2y + 2 = | vedclass

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

Question

(D) Given equation of the parabola is $y^2 - x - 2y + 2 = 0$.Rearranging the terms,we get $y^2 - 2y = x - 2$.Completing the square on the left side: $

Vedclass · Accepted Answer

$(5/4, 1)$

Vedclass · Answer

(D) Given equation of the parabola is $y^2 - x - 2y + 2 = 0$.Rearranging the terms,we get $y^2 - 2y = x - 2$.Completing the square on the left side: $(y - 1)^2 - 1 = x - 2$.$(y - 1)^2 = x - 1$.Let $Y = y - 1$ and $X = x - 1$. The equation becomes $Y^2 = X$.Comparing this with the standard form $Y^2 = 4aX$,we have $4a = 1$,so $a = 1/4$.The focus of $Y^2 = 4aX$ is $(a, 0)$,which is $(1/4, 0)$.To find the focus in the $(x, y)$ coordinate system,we substitute back:$X = 1/4 \implies x - 1 = 1/4 \implies x = 5/4$.$Y = 0 \implies y - 1 = 0 \implies y = 1$.Thus,the focus is $(5/4, 1)$.

The focus of the parabola $y^2 - x - 2y + 2 = 0$ is

Similar Questions

If $a \ne 0$ and the line $2bx + 3cy + 4d = 0$ passes through the points of intersection of the parabolas $y^2 = 4ax$ and $x^2 = 4ay$,then

Normals at $P$,$Q$,and $R$ are drawn to the parabola $y^2 = 4x$ which intersect at the point $(3, 0)$. Then,the triangle $\Delta PQR$ is:

The coordinates of the ends of a focal chord of a parabola $y^2 = 4ax$ are $(x_1, y_1)$ and $(x_2, y_2)$. Then $x_1x_2 + y_1y_2$ has the value equal to

Let $P$ be the parabola,whose focus is $(-2, 1)$ and directrix is $2x + y + 2 = 0$. Then the sum of the ordinates of the points on $P$,whose abscissa is $-2$,is

The number of normals drawn to the parabola $y^2=4x$ from the point $(1,0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module