The graph of the curve `x^2+y^2-2x y-8x-8y+32=0` falls wholly in the (a) first quadrant (b) second quadrant (c) third quadrant (d) none of these

Options

A. first quadrant
B. second quadrant
C. third quadrant
D. none of these

Answer

A

Explanation

(1) `x^(2)+y^(2)-2xy-8x-8y+32=0`
`or(x-y)^(2)=8(x+y-4)`
is a parabola whose axis is x-y=0 and the tangent at the vertex is x+y-4=0.
Also, when y=0., we have `x^(2)-8x+32=0` which gives no real values of x.
When x=0, we have `y^(2)-8x+32=0` which gives no real values of y.
So, the parabola does not intersect the axes. Hence, the graph falls in the first quadrant.