If one of the lines given by the equation 2x^ | vedclass

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

Question

(D) Let the slopes of the lines represented by $2x^2 + axy + 3y^2 = 0$ be $m$ and $m_1$. Then $m + m_1 = -a/3$ and $mm_1 = 2/3$. Let the slopes of the

Vedclass · Accepted Answer

$a = -5$ and $b = -1$

Vedclass · Answer

(D) Let the slopes of the lines represented by $2x^2 + axy + 3y^2 = 0$ be $m$ and $m_1$. Then $m + m_1 = -a/3$ and $mm_1 = 2/3$. Let the slopes of the lines represented by $2x^2 + bxy - 3y^2 = 0$ be $m$ and $m_2$. Then $m + m_2 = -b/(-3) = b/3$ and $mm_2 = 2/(-3) = -2/3$. Since the other two lines are perpendicular,$m_1m_2 = -1$. Dividing the product of slopes: $(mm_1) / (mm_2) = (2/3) / (-2/3) = -1$. Thus,$m_1 / m_2 = -1$,which implies $m_1 = -m_2$ or $m_1 + m_2 = 0$. From $mm_1 = 2/3$ and $mm_2 = -2/3$,we have $m_1 = 2/(3m)$ and $m_2 = -2/(3m)$. Substituting into $m_1m_2 = -1$: $(2/(3m)) 	imes (-2/(3m)) = -1$ $\Rightarrow -4/(9m^2) = -1$ $\Rightarrow m^2 = 4/9$ $\Rightarrow m = 2/3$ (taking $m = 2/3$). Then $m_1 = (2/3) / (2/3) = 1$ and $m_2 = (-2/3) / (2/3) = -1$. Using $m + m_1 = -a/3$: $2/3 + 1 = -a/3$ $\Rightarrow 5/3 = -a/3$ $\Rightarrow a = -5$. Using $m + m_2 = b/3$: $2/3 - 1 = b/3$ $\Rightarrow -1/3 = b/3$ $\Rightarrow b = -1$.

If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincides with one of those given by the equation $2x^2 + bxy - 3y^2 = 0$,while the other two lines are perpendicular to each other,then the values of $a$ and $b$ are

Similar Questions

$A$ pair of lines $S=0$ together with the lines given by the equation $8 x^2-14 x y+3 y^2+10 x+10 y-25=0$ form a parallelogram. If its diagonals intersect at the point $(3,2)$,then the equation $S=0$ is

If two pairs of straight lines with combined equations $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ form a square,then the combined equation of its diagonals is

If the lines represented by $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ intersect on the $x$-axis,which of the following is in general incorrect?

The numbers $\alpha$ and $\beta$ are such that one of the lines of $2x^2 + \alpha xy + 3y^2 = 0$ coincides with one of the lines of $2x^2 + \beta xy - 3y^2 = 0$. If the two lines other than that common line are perpendicular,then $|\alpha + \beta|$ is equal to

The equations to a pair of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations to its diagonals are

Vedclass Test Series

Exam Paper Generator

Online Exam Module