Answer to Question #91094 in Linear Algebra for Kamlesh Devi

Question #91094
Reduce the following equation to standard form hence identify the conic it represents x2 - 3xy + y2 + 4x - 4y - 5 = 0
1
Expert's answer
2019-06-24T10:16:06-0400

equation:


"x^2-3xy+y^2+4x-4y-5=0"

"Ax^2+Bxy+Cy^2+Dx+Ey+F=0"

Find eigenvalues of matrix


"A_{33}=\\begin{pmatrix}\n A & B\/2 \\\\\n B\/2 & C\n\\end{pmatrix}"

from equation


"\\begin{vmatrix}\n A-z & B\/2 \\\\\n B\/2 & C-z\n\\end{vmatrix}=0"

substituting A = 1, B = -3, C = 1 in this equation


"\\begin{vmatrix}\n 1-z & -3\/2 \\\\\n -3\/2 & 1-z\n\\end{vmatrix}=0"

"(1-z)^2-(-3\/2)^2=0"

"(1-3\/2-z)(1+3\/2-z)=0"

"(z+1\/2)(z-5\/2)=0"

"z_1=-1\/2, z_2=5\/2"

Let


"A_q=\\begin{pmatrix}\n A& B\/2 &D\/2\\\\\n B\/2 & C&E\/2\\\\\n D\/2&E\/2&F\n\\end{pmatrix}=\n\\begin{pmatrix}\n 1& -3\/2 &2\\\\\n -3\/2 & 1&-2\\\\\n 2&-2&-5\n\\end{pmatrix}"

Standard form of equation


"x^2-3xy+y^2+4x-4y-5=0"

can be found using formula:


"z_1x'^2+z_2y'^2=-detA_q\/detA_{33}"

"detA_{33}=\\begin{vmatrix}\n 1& -3\/2 \\\\\n -3\/2 & 1\n\\end{vmatrix}=1-9\/4=-5\/4"

"detA_q=\\begin{vmatrix}\n 1& -3\/2 &2\\\\\n -3\/2 & 1&-2\\\\\n 2&-2&-5\n\\end{vmatrix}="

add the first row to the second

"\\begin{vmatrix}\n 1& -3\/2 &2\\\\\n -1\/2 & -1\/2&0\\\\\n 2&-2&-5\n\\end{vmatrix}="

subtract the first column from the second

"\\begin{vmatrix}\n 1& -5\/2 &2\\\\\n -1\/2 & 0&0\\\\\n 2&-4&-5\n\\end{vmatrix}="

"-(-1\/2)(-5\/2*(-5)-(-4)*2)=41\/4"

Substitute all found values in the equation


"z_1x'^2+z_2y'^2=-detA_q\/detA_{33}"

"-1\/2x'^2+5\/2y'^2=-(41\/4)\/(-5\/4)=41\/5"

dividing by (41/5)


"-1\/2\/(41\/5)x'^2+5\/2\/(41\/5)y'^2=1"

"-x'^2\/(82\/5)+y'^2\/(82\/25)=1"

since coefficients of squared terms have different signs, this is standard equation of hyperbola.

Answer: standard equation:

"-x'^2\/(82\/5)+y'^2\/(82\/25)=1"

conic section is hyperbola.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS