If the slope of one of the lines given by ax^ | vedclass

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

Question

(B) Let the slopes of the two lines be $m_1$ and $m_2$. Given $m_2 = 5m_1$.From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{

Vedclass · Accepted Answer

$5h^2 = 9ab$

Vedclass · Answer

(B) Let the slopes of the two lines be $m_1$ and $m_2$. Given $m_2 = 5m_1$.From 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 $m_2 = 5m_1$ into the sum: $m_1 + 5m_1 = 6m_1 = -\frac{2h}{b} \implies m_1 = -\frac{h}{3b}$.Substituting $m_2 = 5m_1$ into the product: $m_1(5m_1) = 5m_1^2 = \frac{a}{b} \implies m_1^2 = \frac{a}{5b}$.Equating the square of the expression for $m_1$: $(-\frac{h}{3b})^2 = \frac{a}{5b} \implies \frac{h^2}{9b^2} = \frac{a}{5b}$.Multiplying both sides by $45b^2$: $5h^2 = 9ab$.

If the slope of one of the lines given by $ax^2 + 2hxy + by^2 = 0$ is $5$ times the other,then

Similar Questions

The equation of the pair of lines passing through the point $(2,1)$ and perpendicular to the pair of lines $4xy + 2x + 6y + 3 = 0$ is:

The equation of the pair of straight lines perpendicular to the pair $2 x^2+3 x y+2 y^2+10 x+5 y=0$ and passing through the origin is $..........$

If the lines represented by the equation $6x^2 + 41xy - 7y^2 = 0$ make angles $\alpha$ and $\beta$ with the $x$-axis,then $\tan \alpha \cdot \tan \beta = $

The length of the $x$-intercept made by the pair of lines $2x^2 + xy - 6y^2 - 2x + 17y - 12 = 0$ is

The joint equation of the pair of lines passing through the origin with slopes $(1+\sqrt{2})$ and $\frac{1}{(1+\sqrt{2})}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module