Suppose the slopes m_1 and m_2 of the lines r | vedclass

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

Question

(A) Given the equation $ax^2+2hxy+by^2=0$,the sum of slopes is $m_1+m_2 = -\frac{2h}{b}$ and the product of slopes is $m_1m_2 = \frac{a}{b}$.From $m_1

Vedclass · Accepted Answer

$\frac{a}{12}=\frac{b}{6}=\frac{h}{\pm 11}$

Vedclass · Answer

(A) Given the equation $ax^2+2hxy+by^2=0$,the sum of slopes is $m_1+m_2 = -\frac{2h}{b}$ and the product of slopes is $m_1m_2 = \frac{a}{b}$.From $m_1m_2-2=0$,we have $\frac{a}{b}=2$,so $a=2b$.From $3(m_1-m_2)-7=0$,we have $m_1-m_2 = \frac{7}{3}$.Using $(m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2$,we get $(\frac{7}{3})^2 = (-\frac{2h}{b})^2 - 4(\frac{a}{b})$.Substituting $a=2b$,we get $\frac{49}{9} = \frac{4h^2}{b^2} - 4(2) = \frac{4h^2}{b^2} - 8$.$\frac{4h^2}{b^2} = \frac{49}{9} + 8 = \frac{49+72}{9} = \frac{121}{9}$.Taking the square root,$\frac{2h}{b} = \pm \frac{11}{3}$,which implies $\frac{h}{b} = \pm \frac{11}{6}$.Thus,$\frac{h}{\pm 11} = \frac{b}{6}$.Since $a=2b$,we have $\frac{a}{2} = b$,so $\frac{a}{12} = \frac{b}{6}$.Therefore,$\frac{a}{12} = \frac{b}{6} = \frac{h}{\pm 11}$.

Suppose the slopes $m_1$ and $m_2$ of the lines represented by $ax^2+2hxy+by^2=0$ satisfy $3(m_1-m_2)-7=0$ and $m_1m_2-2=0$. Then,which of the following is true?

Similar Questions

The area (in square units) of the triangle formed by the line $x+y+1=0$ and the pair of straight lines $x^2-3xy+2y^2=0$ is

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

The product of the lengths of the perpendiculars drawn from the point $(-1, 5)$ to the pair of lines $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$ is

The orthocentre of the triangle formed by the lines $x+y=1$ and $2y^2-xy-6x^2=0$ is

If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of lines $3x^2+11xy-4y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+12=0$,then $2(a-h+b-g+f-12)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module