The image of the pair of lines represented by | vedclass

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

Question

(D) The given equation is $ax^2 + 2hxy + by^2 = 0$. Let the lines be $y = m_1x$ and $y = m_2x$. Then $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a

Vedclass · Accepted Answer

$ax^2 - 2hxy + by^2 = 0$

Vedclass · Answer

(D) The given equation is $ax^2 + 2hxy + by^2 = 0$. Let the lines be $y = m_1x$ and $y = m_2x$. Then $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$. The image of a line $y = mx$ in the mirror $y = 0$ (the $x$-axis) is $y = -mx$. Thus,the new lines are $y = -m_1x$ and $y = -m_2x$. The combined equation is $(y + m_1x)(y + m_2x) = 0$. Expanding this,we get $y^2 + (m_1 + m_2)xy + m_1m_2x^2 = 0$. Substituting the values,we get $y^2 + (-\frac{2h}{b})xy + (\frac{a}{b})x^2 = 0$. Multiplying by $b$,we get $by^2 - 2hxy + ax^2 = 0$,which is $ax^2 - 2hxy + by^2 = 0$.

The image of the pair of lines represented by $ax^2 + 2hxy + by^2 = 0$ by the line mirror $y = 0$ is

Similar Questions

In $\triangle ABC$,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

The Cartesian form of the polar equation $\theta = \tan^{-1} 2$ is

The solution of the differential equation $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$ is

The type of bond connecting two nucleotides is

Velocity and acceleration vectors of a charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$ respectively. Then the value of $c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module