If the ratio of gradients of the lines repres | vedclass

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

Question

(C) Let the gradients of the lines be $m_1$ and $m_2$. Given $m_1 : m_2 = 1 : 3$,so let $m_1 = m$ and $m_2 = 3m$.For the equation $ax^2 + 2hxy + by^2

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(C) Let the gradients of the lines be $m_1$ and $m_2$. Given $m_1 : m_2 = 1 : 3$,so let $m_1 = m$ and $m_2 = 3m$.For the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$.Substituting the values,$m + 3m = 4m = -\frac{2h}{b}$,which gives $m = -\frac{h}{2b}$.Also,$m(3m) = 3m^2 = \frac{a}{b}$.Substituting $m = -\frac{h}{2b}$ into the second equation: $3(-\frac{h}{2b})^2 = \frac{a}{b}$.$3(\frac{h^2}{4b^2}) = \frac{a}{b}$.$\frac{3h^2}{4b^2} = \frac{a}{b} \Rightarrow \frac{h^2}{ab} = \frac{4b^2}{3b^2} = \frac{4}{3}$.Alternatively,for a ratio $1 : n$,the condition is $\frac{h^2}{ab} = \frac{(n+1)^2}{4n}$.For $n = 3$,$\frac{h^2}{ab} = \frac{(3+1)^2}{4(3)} = \frac{16}{12} = \frac{4}{3}$.

If the ratio of gradients of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is $1 : 3$,then the value of the ratio $h^2 : ab$ is

Similar Questions

If $h^2=ab$,then the slopes of lines represented by $ax^2+2hxy+by^2=0$ are in the ratio

The equation $\frac{x^2}{a} + \frac{xy}{h} + \frac{y^2}{b} = 0$ $(a \neq 0, h \neq 0, b \neq 0)$ represents two coincident lines if:

If $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$ represents a pair of straight lines such that the slope of one of the lines is twice the other,then $\frac{ab}{h^2} =$

If the lines represented by the equation $2x^2 - pxy + 2y^2 = 0$ are real,then the value of '$p$' lies in the interval

If one of the lines given by the pair of lines $3x^2 + axy - 2y^2 = 0$ makes an angle of $60^{\circ}$ with the $x$-axis,then $a=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module