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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Focus of the parabola
$x^2 + y^2 - 2xy - 4(x + y - 1) = 0$ is:
Solution
Given equation: $x^2 + y^2 - 2xy - 4x - 4y + 4 = 0$
Group terms:
$(x - y)^2 - 4(x + y) + 4 = 0$
Let:
$u = x - y,\;\; v = x + y$
Then equation becomes:
$u^2 - 4v + 4 = 0$
$u^2 = 4(v - 1)$
This is the standard parabola:
$u^2 = 4p(v - 1)$
Comparing gives:
$4p = 4 \Rightarrow p = 1$
Vertex in $(u,v)$:
$(0,1)$
Focus in $(u,v)$:
$(0, 1 + p) = (0,2)$
Convert to $(x,y)$:
$x - y = 0$
$x + y = 2$
Solving:
$x = 1,\; y = 1$
Therefore, the focus is:
$\boxed{(1,1)}$
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