The equation x^2-y^2+ax+b=0 represents a pair | vedclass

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

Question

(D) The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. This represents a pair of straight lines if the determi

Vedclass · Accepted Answer

$6, 9$

Vedclass · Answer

(D) The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. This represents a pair of straight lines if the determinant $\Delta = \begin{vmatrix} A & H & G \ H & B & F \ G & F & C \end{vmatrix} = 0$. Comparing $x^2 - y^2 + ax + b = 0$ with the general form,we have $A=1, B=-1, C=b, H=0, G=\frac{a}{2}, F=0$. Substituting these values into the determinant condition: $\begin{vmatrix} 1 & 0 & \frac{a}{2} \ 0 & -1 & 0 \ \frac{a}{2} & 0 & b \end{vmatrix} = 0$. Expanding along the second row: $-1 	imes (1 	imes b - \frac{a}{2} 	imes \frac{a}{2}) = 0$. $-1 	imes (b - \frac{a^2}{4}) = 0$ $\Rightarrow b - \frac{a^2}{4} = 0$ $\Rightarrow a^2 = 4b$. Checking the options: For $(6, 9)$,$6^2 = 36$ and $4 	imes 9 = 36$. Thus,the ordered pair $(6, 9)$ satisfies the condition.

The equation $x^2-y^2+ax+b=0$ represents a pair of lines for the ordered pair $(a, b) =$

Similar Questions

If the sum of the slopes of the lines given by ${x^2} - 2cxy - 7{y^2} = 0$ is four times their product,then $c$ has the value

The joint equation of two lines passing through the origin,each making an angle of $30^{\circ}$ with the positive $Y$-axis,is

If one of the lines given by $k x^{2}+x y-y^{2}=0$ bisects the angle between the coordinate axes,then the values of $k$ are

The joint equation of the pair of lines passing through $(3, -2)$ and parallel to the lines represented by $x^{2} - 4xy + 3y^{2} = 0$ is:

The equation of the line common to the pair of lines $(p^2-q^2) x^2+(q^2-r^2) xy+(r^2-p^2) y^2=0$ and $(l-m) x^2+(m-n) xy+(n-l) y^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module