The graph of the curve `x^2+y^2-2x y-8x-8y+32=0` falls wholly in the first quadrant (b) second quadrant 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.