The equation of potato chips is
\(z=\frac{x^2}{a^2}-\frac{y^2}{a^2} \) ......\(1\)
Here, \(a\) is constant.
Consider that the height of the chips above the z-axis is \(z_{1}\) and the height below the z-axis is \(z_{2}\).
Since the height of the chips is \(0.16\) inch, therefore,
\(z_{1}-\left(-z_{2}\right)=0.16\)
\(z_{1}-z_{2}=0.16\)
Take the highest point \(P\) on the chips lies in the xz-plane.
Thus, the coordinates of the point \(P\) is \(\left(1.3, 0, z_{1}\right)\).
Substitute \(x = 1.3,y = 0\), and \(z=z_{1}\) in equation \(\left(1\right)\).
\(z_{1}=\frac{1.3^2}{a^2}-\frac{0^2}{a^2} \)
\(z_{1}=\frac{1.3^2}{a^2}\)
Take the lowest point \(Q\) on the chips lies in the yz-plane.
Thus, the coordinates of the point \(Q\) is \(\left(0,0.8,-z_{2}\right)\)
Substitute \(x=0,y=0.8\), and \(z=-z_{2}\) in equation \(\left(1\right)\).
\(-z_{2}=\frac{0^2}{a^2}-\frac{0.8^2}{a^2} \)
\(0.16-z_{1}=\frac{0.8^2}{a^2} \)
\(z_{1}=0.16-\frac{0.8^2}{a^2} \)
Substitute \(z_{1}=0.16-\frac{0.8^2}{a^2} \) in \(z_{1}=\frac{1.3^2}{a^2} \) .
\(0.16-\frac{1.3^2}{a^2}=\frac{0.8^2}{a^2}\)
\(a^2=\frac{1.3^2+0.8^2}{0.16}\)
\(a=\pm\sqrt{14.5625}\)
\(\approx\pm3.81\)
Hence the value of \(a\) is \(\pm3.81\) .
Substitute \(a=\pm3.81\) in equation \(\left(1\right)\).
\(z=\frac{x^2}{3.81^2}-\frac{y^2}{3.81^2} \) .
Hence, the equation of the chips is \(z=\frac{x^2}{3.81^2}-\frac{y^2}{3.81^2} \) .
