Question

The graph of the curve ${{x}^{2}}+{{y}^{2}}-2xy-8x-8y+32=0$ falls wholly in the
(A) First quadrant
(B) Second quadrant
(C) Third quadrant
(D) none

Answer 1

Hint: Transform the equation ${{x}^{2}}+{{y}^{2}}-2xy-8x-8y+32=0$ as ${{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$ . Now, the given equation is the equation of the parabola whose axis is $\left( x-y \right)=0$ and the equation of the tangent at the vertex is $\left( x+y-4 \right)=0$ . Now, plot the graph. The equation of x-axis is $y=0$ and the equation of the y-axis is $x=0$ . For point of intersection of the parabola and x-axis, put $y=0$ in the equation ${{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$ . Similarly, for point of intersection of the parabola and y-axis, put $x=0$ in the equation ${{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$ . Now, solve it further and check whether the parabola touches or intersect the x-axis and y-axis.

Complete step by step solution:
According to the question, it is given that the equation of the curve is,
${{x}^{2}}+{{y}^{2}}-2xy-8x-8y+32=0$ ………………………(1)
From equation (1), we can figure out about the curve. So, first of all, we need to simplify it in order to figure out the curve.
Simplifying equation (1), we get

& {{x}^{2}}+{{y}^{2}}-2xy-8x-8y+32=0 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-2xy=8x+8y-32 \\\ \end{aligned}$$ $$\Rightarrow {{x}^{2}}+{{y}^{2}}-2xy=8\left( x+y-4 \right)$$ …………………….(2) We know the formula, $${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$$ ……………………(3) Now, from equation (2) and equation (3), we get $$\Rightarrow {{x}^{2}}+{{y}^{2}}-2xy=8\left( x+y-4 \right)$$ $$\Rightarrow {{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$$ ………………………..(4) The above equation represents a parabola whose axis is $$\left( x-y \right)=0$$ and the tangent at the vertex is $$\left( x+y-4 \right)=0$$ . The equation of the axis of the parabola is, $$\left( x-y \right)=0$$ $$\Rightarrow x=y$$ ………………………(5) Clearly, we can see that the equation of the axis of the parabola is a straight line with slope equal to 1. The equation of the tangent at the vertex of the parabola is, $$\left( x+y-4 \right)=0$$ $$\Rightarrow x+y=4$$ ……………………..(6) The above equation is of a straight line whose x-intercept and y-intercept are equal to 4. Now, using equation (4), equation (5), and equation (6), plotting the graph of the parabola, we get  For this parabola to be in the second quadrant or third quadrant or fourth quadrant, the parabola should intersect the axis. Let us check where the parabola intersects the x- axis. The equation of x-axis is $$y=0$$ . From equation (4), we have the equation of the parabola. Putting $$y=0$$ in equation (4), we get, $$\Rightarrow {{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$$ $$\begin{aligned} & \Rightarrow {{\left( x-0 \right)}^{2}}=8\left( x+0-4 \right) \\\ & \Rightarrow {{x}^{2}}=8x-32 \\\ & \Rightarrow {{x}^{2}}-8x+32=0 \\\ \end{aligned}$$ The discriminant of the above quadratic equation is -64 which is negative. It means the quadratic equation does not have any real solution. So, we can say that the parabola also has no solution for $$y=0$$ . Therefore, the parabola doesn’t even touch or intersect the x- axis. Similarly, let us check where the parabola intersects the y- axis. The equation of y-axis is $$x=0$$ . From equation (4), we have the equation of the parabola. Putting $$x=0$$ in equation (4), we get, $$\Rightarrow {{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$$ $$\begin{aligned} & \Rightarrow {{\left( 0-y \right)}^{2}}=8\left( 0+y-4 \right) \\\ & \Rightarrow {{y}^{2}}=8y-32 \\\ & \Rightarrow {{y}^{2}}-8y+32=0 \\\ \end{aligned}$$ The discriminant of the above quadratic equation is -64 which is negative. It means the quadratic equation does not have any real solution. So, we can say that the parabola also has no solution for $$x=0$$ . Therefore, the parabola doesn’t even touch or intersect the y- axis. Thus, we can say that the parabola is in the first quadrant. Hence, option (A) is the correct option. Note: In this question, one might get confused about how to get the equation of the axis and the equation of the tangent at the vertex. So, don’t be confused here. This is the equation of the parabola and the equation of the axis of the parabola is given by making the square term equal to zero and the equation of the tangent at the vertex is given by making the linear terms equal to zero. Like, in the equation $${{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$$ , The square term is $$\left( x-y \right)$$ . So, the equation of the axis of the parabola is $$\left( x-y \right)=0$$ . Here, in the equation $${{\left( x-y \right)}^{2}}=8\left( x+y-4 \right)$$ , the linear term is $$\left( x+y-4 \right)$$ . So, the equation of the tangent at the vertex is $$\left( x+y-4 \right)=0$$ .