If the slope of one of the lines represented | vedclass

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

Question

(A) Let the slopes of the lines be $m_1$ and $m_2$. Given $m_1 = m_2^2$.From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$

Vedclass · Accepted Answer

$a^2b + ab^2 - 6abh + 8h^3 = 0$

Vedclass · Answer

(A) Let the slopes of the lines be $m_1$ and $m_2$. Given $m_1 = m_2^2$.From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.Substituting $m_1 = m_2^2$ in the sum and product:$m_2^2 + m_2 = -\frac{2h}{b}$ $(i)$$m_2^2 \cdot m_2 = m_2^3 = \frac{a}{b} \Rightarrow m_2 = (\frac{a}{b})^{1/3}$ $(ii)$Substituting $m_2$ from $(ii)$ into $(i)$:$(\frac{a}{b})^{2/3} + (\frac{a}{b})^{1/3} = -\frac{2h}{b}$Cubing both sides:$(\frac{a}{b})^2 + \frac{a}{b} + 3(\frac{a}{b})^{2/3}(\frac{a}{b})^{1/3} [(\frac{a}{b})^{2/3} + (\frac{a}{b})^{1/3}] = -\frac{8h^3}{b^3}$Using $(i)$,we substitute the term in the bracket:$\frac{a^2}{b^2} + \frac{a}{b} + 3(\frac{a}{b})(-\frac{2h}{b}) = -\frac{8h^3}{b^3}$$\frac{a^2}{b^2} + \frac{a}{b} - \frac{6ah}{b^2} = -\frac{8h^3}{b^3}$Multiplying by $b^3$:$a^2b + ab^2 - 6abh = -8h^3$$a^2b + ab^2 - 6abh + 8h^3 = 0$.

If the slope of one of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is the square of the other,then:

Similar Questions

The lines represented by the equation $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$ and the line $4x - 3y - 2 = 0$:

The value of $\lambda$ such that $\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0$ represents a pair of straight lines is:

If $4ab = 3h^2$,then the ratio of slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is

The lines represented by the equation $a{x^2}(b - c) - xy(a(b - c) + c(a - b)) + c{y^2}(a - b) = 0$ are

If the sum of the slopes of the lines given by ${x^2} - 2cxy - 7{y^2} = 0$ is four times their product,then $c$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module