If the axes are rotated through an angle 45^{ | vedclass

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

Question

(A) Given the rotation angle $	heta = 45^{\circ}$.Let the new coordinates be $(x', y')$ and the old coordinates be $(x, y)$.The transformation equati

Vedclass · Accepted Answer

$(x+y)^2=4\sqrt{2}a(x-y)$

Vedclass · Answer

(A) Given the rotation angle $	heta = 45^{\circ}$.Let the new coordinates be $(x', y')$ and the old coordinates be $(x, y)$.The transformation equations are:$x = x' \cos 45^{\circ} - y' \sin 45^{\circ} = \frac{x'-y'}{\sqrt{2}}$$y = x' \sin 45^{\circ} + y' \cos 45^{\circ} = \frac{x'+y'}{\sqrt{2}}$Substituting these into the equation $y^2 = 4ax$:$(\frac{x'+y'}{\sqrt{2}})^2 = 4a(\frac{x'-y'}{\sqrt{2}})$$\frac{(x'+y')^2}{2} = \frac{4a(x'-y')}{\sqrt{2}}$$(x'+y')^2 = \frac{8a}{\sqrt{2}}(x'-y') = 4\sqrt{2}a(x'-y')$Replacing $(x', y')$ with $(x, y)$,the transformed equation is $(x+y)^2 = 4\sqrt{2}a(x-y)$.

If the axes are rotated through an angle $45^{\circ}$ about the origin in an anticlockwise direction,then the transformed equation of $y^2=4ax$ is

Similar Questions

If the axes are rotated by an angle of $-\pi /3$ in the negative direction and the coordinates of a point in the new system are $(4, 2)$,find the coordinates of the point in the original system.

The point to which the origin should be shifted so that the equation $y^2-6y-4x+13=0$ will not contain any term in $y$ and the constant term,is

Let $L$ be the line $2x + y = 2$. If the axes are rotated by $45^\circ$,then the intercepts made by the line $L$ on the new axes are respectively

If the axes are transformed to the point $(-1, 1)$,then the equation $3x^2 + y^2 + 2x + 4y + 15 = 0$ would transform to:

When the origin is shifted to $(-1, 2)$ by the translation of axes,the transformed equation of $x^2+y^2+2x-4y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module