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Chapter 8: Problem 46

Find a point on the curve \(x^{2}+2 y^{2}=6\) whose distance from the line \(x+y=7\) is minimum.

Short Answer

Expert verified
The coordinates of the point (x,y) on the ellipse which is closest to the line can be found by minimizing the distance between a point on the ellipse and the line. The minimum distance is obtained by differentiating distance function, equating it to zero and checking if second derivative is positive. Actual coordinates can be calculated by substituting the obtained parameter values in step 5 into the parametric form equations of x and y from the ellipse.

Step by step solution

01

Understanding the problem

The given equation is \(x^{2}+2 y^{2}=6\), which represents an ellipse. Additionally, the equation of the line is given as \(x+y=7\). The target is to find the coordinates of the point on the ellipse that lies closest to this line.
02

Rearranging the equation of line

We rearrange the equation of the line in the form \(ax+by+c=0\), as needed for the distance formula. Subtract 'x' and 'y' on both sides leads to \(x+y-7=0\). So, a=1, b=1 and c=-7.
03

Differentiation of Ellipse

The distance between the line and point (x, y) on the ellipse can be minimized on changing x and y with respect to a certain parameter t. To find the parametric form of the ellipse we'll introduce a parameter t, express x and y in terms of t. From the equation of the ellipse, we get \(x= \sqrt{6-2y^{2}}\). Differentiating w.r.t y, we get \(dx/dy = -2y/(\sqrt{6-2y^{2}})\). Denoting dx/dy as \(x'\), and equating \(x' = dy/dx = tan(t)\); we obtain \(tan(t) = -2y/(\sqrt{6-2y^{2}})\). Solving the above equation gives \(y = sqrt{6}/2 * sin(t)\) and \(x = sqrt{6}/2 * cos(t)\). The two equations are parametric equations of the ellipse.
04

Apply Distance formula

The distance d from the line and point (x, y) can be written as \(d = \frac{|x+y-7|}{\sqrt{1^{2}+1^{2}}}\). Substitute x and y values from step 3 into the distance function, we get the distance becomes \(d = \frac{| sqrt{6}/2 * cos(t)+ sqrt{6}/2 * sin(t) -7|}{\sqrt{2}}\).
05

Minimize distance

Differentiate the distance d with respect to t and set it equals to zero. Thus we find the required minimum t. Use this t back in parametric equations of x and y to get the coordinates of the point \((x,y)\).
06

Verify minimum distance

Differentiate the distance once more with respect to t to check if the obtained values from step 5 yield minimum distance. If the second derivative is positive, the distance is minimum.

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