(Linear Algebra)7.4: Applications of EigenvaluesPls help thanks Use the Principal Axes Theorem to perform a rotation of axes to eliminate the xy-term in the quadratic equation.2x2 + 4xy + 2y² + 6V2x + 2/2y + 9 = 0(a) Identify the resulting rotated conic.ellipseparabolaO hyperbola(b) Give its equation in the new coordinate system. (Use xp and yp as the new coordinates.)

Given, 2x2+4xy+2y2+62x+22y+9=0By Principal axes theorem: For a conic whose equation is ax2+bxy+cy2+dx+ey+f=0 the rotation X = PX' eliminates the xy-termwhen P is an orthogonal matrix with |P| = 1, that diagonalize the matrix of the quadratic form A. that is PTAP=D=λ100λ2 where λ1 and λ2 are eigenvalues of A. The equation of the rotated conic is λ1(x')2+λ2(y')2+[de]PX'+f=0 Now, the given quadratic equation is 2x2+4xy+2y2+62x+22y+9=0 So, the matrix of the quadratic form associated with this equation is A=2222For Eigenvalues : det(A-λI)=0⇒ 2-λ222-λ=0⇒ 2-λ2-4=0⇒ 2-λ2=4⇒ 2-λ=±2⇒ λ=2±2∴ λ=0, 4For λ1=0 2-0222-0xy=00⇒ 2222xy=00⇒ 1100xy=00 ∵ By row reduce echelon form⇒ x=-yLet y=1, then v1=-11For λ2=4 2-4222-4xy=00⇒ -222-2xy=00⇒ -1100xy=00 ∵ By row reduce echelon form⇒ x=yLet y=1, then v2=11Now the matrix whose columns correspond to these normalized eigenvectors is P=-12121212 Where P is an orthogonal matrix. PTAP=D=0004 Which is a diagonal matrix. ∴ The equation of rotated conic is given by λ1(x')2+λ2(y')2+dePX'+f=0Here, λ1=0, λ2=4, d=62, e=22, and f=9So, (0)(x')2+4(y')2+6222-12121212x'y,+9=0⇒ 4(y')2+6222-12121212x'y,+9=0Here, 6222-12121212x'y,=-48x'y, =-4x'+8y'∴ 4(y')2-4x'+8y'+9=0Replace x' by xp and y' by yp, then we have 4yp2-4xp+8yp+9=0⇒ 4yp2+2yp+1-1-4xp+9=0⇒ 4yp2+2yp+1-4-4xp+9=0⇒ 4yp+12-4xp+5=0⇒ -4xp=-4yp+12-5∴ xp=(yp+1)2+54Which is the required parabola equation in the new coordinate system.