When the origin is shifted to the point left( | vedclass

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

Question

(D) Substituting $x=X+\frac{3}{2}$ and $y=Y+\frac{3}{2}$ into the equation $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$: $32(X+\frac{3}{2})^2 + 8(X+\frac{3}

Vedclass · Accepted Answer

$32 X^2+8 X Y+32 Y^2-63=0$

Vedclass · Answer

(D) Substituting $x=X+\frac{3}{2}$ and $y=Y+\frac{3}{2}$ into the equation $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$: $32(X+\frac{3}{2})^2 + 8(X+\frac{3}{2})(Y+\frac{3}{2}) + 32(Y+\frac{3}{2})^2 - 108(X+\frac{3}{2}) - 108(Y+\frac{3}{2}) + 99 = 0$ Expanding the terms: $32(X^2 + 3X + \frac{9}{4}) + 8(XY + \frac{3}{2}X + \frac{3}{2}Y + \frac{9}{4}) + 32(Y^2 + 3Y + \frac{9}{4}) - 108X - 162 - 108Y - 162 + 99 = 0$ $32X^2 + 96X + 72 + 8XY + 12X + 12Y + 18 + 32Y^2 + 96Y + 72 - 108X - 108Y - 162 - 162 + 99 = 0$ Combining like terms: $32X^2 + 32Y^2 + 8XY + (96+12-108)X + (96+12-108)Y + (72+18+72-162-162+99) = 0$ $32X^2 + 8XY + 32Y^2 - 63 = 0$

When the origin is shifted to the point $\left(\frac{3}{2}, \frac{3}{2}\right)$ by the translation of coordinate axes,then the transformed equation of $32 x^2+8 x y+32 y^2-108 x-108 y+99=0$ is

Similar Questions

If the equation $3x^2 + 4y^2 - xy + k = 0$ is the transformed equation of $3x^2 + 4y^2 - xy - 5x - 7y + 2 = 0$ after shifting the origin to the point $(\alpha, \beta)$ by the translation of axes,then $\alpha + \beta - k =$

Suppose the axes are to be rotated through an angle $\theta$ so as to remove the $xy$ term from the equation $3x^2+2\sqrt{3}xy+y^2=0$. Then in the new coordinate system,the equation $x^2+y^2+2xy=2$ is transformed to:

The angle through which the coordinate axes are to be rotated to remove the $xy$ term in the equation $x^2+2xy-y^2=0$ is

The transformed equation of $x^2+y^2=r^2$ when the axes are rotated through an angle $36^{\circ}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module