Find the centre of the conic section represen | vedclass

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

Question

(A) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$,we get:$

Vedclass · Accepted Answer

$(2, 3)$

Vedclass · Answer

(A) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing this with $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$,we get:$a = 14, h = -2, b = 11, g = -22, f = -29, c = 71$.The centre $(x, y)$ of the conic is given by the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.$\frac{\partial}{\partial x} = 28x - 4y - 44 = 0 \implies 7x - y = 11$  $(1)$$\frac{\partial}{\partial y} = -4x + 22y - 58 = 0 \implies -2x + 11y = 29$  $(2)$Multiplying $(1)$ by $11$,we get $77x - 11y = 121$.Adding this to $(2)$,we get $75x = 150$,so $x = 2$.Substituting $x = 2$ in $(1)$,$7(2) - y = 11 \implies 14 - y = 11 \implies y = 3$.Thus,the centre is $(2, 3)$.

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

Similar Questions

If both roots of the equation $x^2 - 5ax + 6a = 0$ exceed $1$,then the range of '$a$' is

To remove the $xy$ term from the second-degree equation $5x^2 + 8xy + 5y^2 + 3x + 2y + 5 = 0$,the coordinate axes are rotated through an angle $\theta$. Then $\theta$ equals:

To convert the equation $2x^2 + 4xy + 5y^2 - 4x - 22y + 29 = 0$ to homogeneous form,the origin is shifted to the point:

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:

For the curve $x^{2}+4xy+8y^{2}=64$, the tangents are parallel to the $x$-axis only at the points

Vedclass Test Series

Exam Paper Generator

Online Exam Module