The value of p for which the equation x^2+pxy | vedclass

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

Question

(A) The general equation of the second degree $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines if $\Delta = abc+2fgh-af^2-bg^2-ch^2 =

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(A) The general equation of the second degree $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines if $\Delta = abc+2fgh-af^2-bg^2-ch^2 = 0$.Comparing the given equation $x^2+pxy+y^2-5x-7y+6=0$ with the general form:$a=1, b=1, c=6, h=\frac{p}{2}, g=-\frac{5}{2}, f=-\frac{7}{2}$.Substituting these values into the condition $\Delta=0$:$(1)(1)(6) + 2(-\frac{7}{2})(-\frac{5}{2})(\frac{p}{2}) - 1(-\frac{7}{2})^2 - 1(-\frac{5}{2})^2 - 6(\frac{p}{2})^2 = 0$$6 + \frac{35p}{4} - \frac{49}{4} - \frac{25}{4} - \frac{6p^2}{4} = 0$Multiply by $4$:$24 + 35p - 49 - 25 - 6p^2 = 0$$-6p^2 + 35p - 50 = 0$$6p^2 - 35p + 50 = 0$$(2p-5)(3p-10) = 0$Thus,$p = \frac{5}{2}$ or $p = \frac{10}{3}$.Given the options,the correct value is $\frac{5}{2}$.

The value of $p$ for which the equation $x^2+pxy+y^2-5x-7y+6=0$ represents a pair of straight lines is:

Similar Questions

The equation of a pair of lines is given by a second-degree homogeneous equation. If one of them is perpendicular to the line $x+2y+7=0$ and the other is parallel to the line $3x+4y+5=0$,then the equation of that pair of lines is:

The equation $x^2 + ky^2 + 4xy = 0$ represents two coincident lines,if $k =$

If the equation $x^2 + y^2 + 2gx + 2fy + 1 = 0$ represents a pair of lines,then

If one of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is coincident with one of the lines represented by $a'x^2 + 2h'xy + b'y^2 = 0$,then:

The two lines represented by the equation $x^2 + xy + y^2 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module