(a, b) is the point to which the origin has t | vedclass

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

Question

(B) Step $1$: Shifting the origin to $(h, k)$ to remove first-degree terms from $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$. Let $x = X + h$ and $y = Y + k$. Sub

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Step $1$: Shifting the origin to $(h, k)$ to remove first-degree terms from $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$. Let $x = X + h$ and $y = Y + k$. Substituting these into the equation,the coefficients of $X$ and $Y$ must be zero. $4h - 3k = 0$ and $-3h + 8k + 5 = 0$. Solving these,we get $h = -\frac{15}{23}$ and $k = -\frac{20}{23}$. The origin is shifted to $(h, k)$,so the point $(a, b)$ is $(-\frac{15}{23}, -\frac{20}{23})$. Step $2$: Substitute $a = -\frac{15}{23}$ and $b = -\frac{20}{23}$ into the equation $ax^2 + 23abxy + by^2 = 0$. $-\frac{15}{23}x^2 + 23(-\frac{15}{23})(-\frac{20}{23})xy - \frac{20}{23}y^2 = 0$. Multiplying by $-23$,we get $15x^2 - 300xy + 20y^2 = 0$,or $3x^2 - 60xy + 4y^2 = 0$. Step $3$: To remove the $xy$-term by rotating axes by angle $	heta$,the formula is $	an 2	heta = \frac{B}{A - C}$,where the equation is $Ax^2 + Bxy + Cy^2 = 0$. Here $A = 3, B = -60, C = 4$. $	an 2	heta = \frac{-60}{3 - 4} = \frac{-60}{-1} = 60$.

Similar Questions

When the origin is shifted to the point $(2, b)$ by translation of axes,the coordinates of the point $(a, 4)$ change to $(6, 8)$. When the origin is shifted to $(a, b)$ by translation of axes,if the transformed equation of $x^2+4xy+y^2=0$ is $X^2+2HXY+Y^2+2GX+2FY+C=0$,then $2H(G+F)=$

The transformed equation of $x^2+6xy+8y^2=10$ when the axes are rotated through an angle $\frac{\pi}{4}$ is:

When the coordinate axes are rotated through an angle $\theta$ in anti-clockwise direction,if the transformed equation of $x^2+y^2+2xy+2x+6y+1=0$ is $(2+\sqrt{3})X^2+2XY+(2-\sqrt{3})Y^2+aX+bY+2=0$,then $3a-b=$

Suppose the axes $X$ and $Y$ are obtained by rotating the axes $x$ and $y$ by an angle $\theta$. If the equation $x^2+2 \sqrt{3} x y-y^2=4 a^2$ is transformed to $X^2-Y^2=2 a^2$ with respect to the $XY$-axes,then $\theta$ is equal to (in $^{\circ}$)

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 $(a+2h+b-\sqrt{c})^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module