Question

Find the focus of the parabola y^2-x-2y+2=0

Answer 1

y^2 - x - 2y + 2 = 0

y^2 - 2y = x - 2

(y - 1)^2 - 1 = x - 2

(y - 1)^2 = x - 1

Comparing with standard equation :-

y^2 = 4ax

Vertex = 0,0

Focus = a, 0

(y - y1) ^2 = 4a(x - x1)

Vertex = x1, y1

Focus = a + x1, y1

(y - 1)^2 = x - 1

(y - 1)^2 = 4(1/4)(x - 1)

here a = 1/4

Vertex = 1,1

Focus = 1 + 1/4, 1

= 5/4, 1

Answered by an IIT JEE ASPIRANT and all India mathematics OLYMPIAD TOPPER in class 10th

Answer 2

$\large\underline\mathrm\red{Solution}$

y² - x - 2y + 2 = 0

y² - 2y + 1 = x - 1

(y - 1)² = x - 1

y² = x

y² = 4. (1/4 x)

$\large\underline\mathrm\red{Now}$, Y = y - 1 and X = x - 1

It's focus is given by (a, 0) = (1/4, 0)

$\large\underline\mathrm\red{Then}$,

$\implies$x = 1/4

$\implies$x - 1 = 1/4

$\implies$x = 5/4

$\implies$y = o

$\implies$y - 1 = 0

$\implies$y = 1

$\large\underline\mathrm\red{Hence}$, focus of hyperbola is (4/5, 1).

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