To eliminate the xy term from the second-degr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation is $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$.Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy

Vedclass · Accepted Answer

$\pi /4$

Vedclass · Answer

(B) The given equation is $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$.Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 5$,$2h = 8$ (so $h = 4$),and $b = 5$.The angle of rotation $	heta$ required to eliminate the $xy$ term is given by the formula $	an(2	heta) = \frac{2h}{a - b}$.Substituting the values,we have $	an(2	heta) = \frac{8}{5 - 5} = \frac{8}{0}$,which is undefined.Therefore,$2	heta = \frac{\pi}{2}$,which implies $	heta = \frac{\pi}{4}$.

To eliminate the $xy$ term from the second-degree equation $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$,the axes are rotated by an angle $\theta$. What is the value of $\theta$?

Similar Questions

In order to eliminate the first degree terms from the equation $2x^2+4xy+5y^2-4x-22y+7=0$,the point to which the origin is to be shifted is:

The point to which the origin is to be shifted to remove the first degree terms from the equation $2x^2+4xy-6y^2+2x+8y+1=0$ is

The equation $\frac{x^2}{1-k} - \frac{y^2}{1+k} = 1$,where $k > 1$,represents:

The equation $16 x^2+y^2+8 x y-74 x-78 y+212=0$ represents

What type of conic section is represented by the equation $13x^2 - 18xy + 37y^2 + 2x + 14y - 2 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module