If the equation 12x^2 - 10xy + 2y^2 + 11x - 5 | vedclass

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

Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Comparing the given equation $12x^2 - 10xy + 2y^2 + 11x - 5y + k = 0$ with the general form:$a = 12, b = 2, c = k, 2h = -10 \implies h = -5, 2g = 11 \implies g = 11/2, 2f = -5 \implies f = -5/2$.Substituting these values into the condition:$(12)(2)(k) + 2(-5/2)(11/2)(-5) - 12(-5/2)^2 - 2(11/2)^2 - k(-5)^2 = 0$$24k + 137.5 - 12(25/4) - 2(121/4) - 25k = 0$$24k - 25k + 137.5 - 75 - 60.5 = 0$$-k + 2 = 0 \implies k = 2$.

If the equation $12x^2 - 10xy + 2y^2 + 11x - 5y + k = 0$ represents two straight lines,then the value of $k$ is

Similar Questions

For $\ell \in R$,the equation $(2 \ell-3) x^2+2 \ell xy-y^2=0$ represents a pair of distinct lines

The joint equation of the straight lines $x + y = 1$ and $x - y = 4$ is

The slopes of the lines represented by $6x^2 + 2hxy + y^2 = 0$ are in the ratio $2:3$,then $h =$

The area of the triangle formed by the pair of lines $23x^2 - 48xy + 3y^2 = 0$ with the line $2x + 3y + 5 = 0$ is

The equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of straight lines $3x^2-8xy+5y^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module