In order to eliminate the first degree terms | vedclass

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

Question

(C) The given equation is $2x^2+4xy+5y^2-4x-22y+7=0$. Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2,

Vedclass · Accepted Answer

$(-2, 3)$

Vedclass · Answer

(C) The given equation is $2x^2+4xy+5y^2-4x-22y+7=0$. Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=2, h=2, b=5, g=-2, f=-11, c=7$. To eliminate the first-degree terms,the origin $(0,0)$ must be shifted to the point $(h', k')$ given by the intersection of the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$. $\frac{\partial}{\partial x} = 4x+4y-4 = 0 \implies x+y=1$ $\frac{\partial}{\partial y} = 4x+10y-22 = 0 \implies 2x+5y=11$ Solving these equations: From the first equation,$x = 1-y$. Substituting into the second: $2(1-y)+5y=11 \implies 2-2y+5y=11 \implies 3y=9 \implies y=3$. Then $x = 1-3 = -2$. Thus,the origin should be shifted to $(-2, 3)$.

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:

Similar Questions

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

When does the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represent an ellipse?

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 $\theta$?

Find the centre of the conic section represented by the equation $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$.

The centre of the conic represented by the equation $2x^2 - 72xy + 23y^2 - 4x - 28y - 48 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module