If 6x^2 + 11xy - 10y^2 + x + 31y + k = 0 repr | vedclass

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

Question

(A) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\De

Vedclass · Accepted Answer

$-15$

Vedclass · Answer

(A) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Comparing the given equation $6x^2 + 11xy - 10y^2 + x + 31y + k = 0$ with the general form:$a = 6, b = -10, c = k, h = \frac{11}{2}, g = \frac{1}{2}, f = \frac{31}{2}$.Substituting these values into the condition $\Delta = 0$:$6(-10)(k) + 2(\frac{31}{2})(\frac{1}{2})(\frac{11}{2}) - 6(\frac{31}{2})^2 - (-10)(\frac{1}{2})^2 - k(\frac{11}{2})^2 = 0$$-60k + \frac{341}{4} - 6(\frac{961}{4}) + \frac{10}{4} - \frac{121k}{4} = 0$Multiply the entire equation by $4$:$-240k + 341 - 5766 + 10 - 121k = 0$$-361k - 5415 = 0$$361k = -5415$$k = -\frac{5415}{361} = -15$.

If $6x^2 + 11xy - 10y^2 + x + 31y + k = 0$ represents a pair of straight lines,then $k = $

Similar Questions

The joint equation of a pair of lines passing through the origin and making an angle of $\frac{\pi}{4}$ with the line $3x + 2y - 8 = 0$ is

The area of the triangle formed by the lines $3x^2 - 4xy + y^2 = 0$ and $2x - y = 6$ is:

The lines ${a^2}{x^2} + bc{y^2} = a(b + c)xy$ will be coincident,if

If a line $L$ is common to the pairs of lines $6 x^2-x y-12 y^2=0$ and $15 x^2+14 x y-8 y^2=0$,then the combined equation of the other two lines is

The equation $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$ will represent two mutually perpendicular straight lines,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module