If theta_1, theta_2, theta_3 are respectively | vedclass

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

Question

(B) To eliminate the $xy$ term in a general quadratic equation $ax^2 + bxy + cy^2 + dx + ey + f = 0$,the axes must be rotated by an angle $	heta$ suc

Vedclass · Accepted Answer

$	heta_3, 	heta_1, 	heta_2$

Vedclass · Answer

(B) To eliminate the $xy$ term in a general quadratic equation $ax^2 + bxy + cy^2 + dx + ey + f = 0$,the axes must be rotated by an angle $	heta$ such that $	an 2	heta = \frac{b}{a-c}$.For $A_1$: $a=3, b=5, c=3$. $	an 2	heta_1 = \frac{5}{3-3} = \infty$ $\Rightarrow 2	heta_1 = \frac{\pi}{2}$ $\Rightarrow 	heta_1 = \frac{\pi}{4}$.For $A_2$: $a=5, b=2\sqrt{3}, c=3$. $	an 2	heta_2 = \frac{2\sqrt{3}}{5-3} = \sqrt{3}$ $\Rightarrow 2	heta_2 = \frac{\pi}{3}$ $\Rightarrow 	heta_2 = \frac{\pi}{6}$.For $A_3$: $a=4, b=\sqrt{3}, c=5$. $	an 2	heta_3 = \frac{\sqrt{3}}{4-5} = -\sqrt{3}$ $\Rightarrow 2	heta_3 = \frac{2\pi}{3}$ $\Rightarrow 	heta_3 = \frac{\pi}{3}$.Comparing the angles: $\frac{\pi}{3} > \frac{\pi}{4} > \frac{\pi}{6}$,which means $	heta_3 > 	heta_1 > 	heta_2$.

Similar Questions

If the area of the region bounded by the curves $y=x^2$ and $x=y^2$ is $k$,then the area of the region bounded by the curves $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$ is:

When the origin is shifted to $(2, 3)$,the transformed equation of a curve becomes $x^2+3xy-2y^2+17x-7y-11=0$. Find the original equation of the curve.

When the coordinate axes are rotated about the origin through an angle $\frac{\pi}{4}$ in the positive direction,the equation $ax^2+2hxy+by^2=c$ is transformed to $25x^2+9y^2=225$,then $(a+2h+b-\sqrt{c})^2=$

When the origin is shifted to the point $P$ by translation of axes,the equation $2x^2+y^2-4x+4y=0$ is transformed to $2x^2+y^2-8x+8y+18=0$. Then the transformed equation of the straight line $x+2y+2=0$ if the origin is shifted to the same point $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module