The combined equation of the pair of lines pa | vedclass

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

Question

(A) The given equation is $2x^2 - xy - y^2 = 0$.Divide by $x^2$ to get the slopes $m = \frac{y}{x}$:$2 - m - m^2 = 0 \implies m^2 + m - 2 = 0$.Let the

Vedclass · Accepted Answer

$x^2 - xy - 2y^2 = 0$

Vedclass · Answer

(A) The given equation is $2x^2 - xy - y^2 = 0$.Divide by $x^2$ to get the slopes $m = \frac{y}{x}$:$2 - m - m^2 = 0 \implies m^2 + m - 2 = 0$.Let the slopes of the given lines be $m_1$ and $m_2$. The slopes of the lines perpendicular to these will be $-\frac{1}{m_1}$ and $-\frac{1}{m_2}$.Replacing $m$ with $-\frac{1}{m}$ in the equation $2 - m - m^2 = 0$:$2 - (-\frac{1}{m}) - (-\frac{1}{m})^2 = 0$$2 + \frac{1}{m} - \frac{1}{m^2} = 0$Multiply by $m^2$:$2m^2 + m - 1 = 0$.Substitute $m = \frac{y}{x}$:$2(\frac{y}{x})^2 + \frac{y}{x} - 1 = 0$$2y^2 + xy - x^2 = 0$Multiplying by $-1$ gives $x^2 - xy - 2y^2 = 0$.

The combined equation of the pair of lines passing through the origin and perpendicular to the pair of lines $2x^2 - xy - y^2 = 0$ is:

Similar Questions

If the two lines given by $ax^2+2hxy+by^2=0$ make inclinations $\alpha$ and $\beta$ with the $x$-axis,then $\tan(\alpha+\beta)=$

If the sum of the slopes of the pair of lines represented by $4x^2 + 2hxy - 7y^2 = 0$ is equal to the product of the slopes,then the value of $h$ is

The area of the triangle formed by the lines $3x^2 - 4xy + y^2 = 0$ and $2x - y = 6$ is:

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module