Question

By shifting the origin to the point (-1,2) through translation of axes, if $ax^2+2hxy+by^2+2gx+2fy+c=0$ is the transformed equation of $2x'^2-x'y'+y'^2-3x'+4y'-5=0$, then $2(f+g+h) =$

• A. $a+b+c$
• B. $a-5(b+c)$
• C. $3(a+b+c)$
• D. $c-5(a+b)$

Hint:

Be very careful with the transformation equations. If the origin is shifted to $(h_0, k_0)$, the substitution is $x_{\text{old}} = x_{\text{new}} + h_0$ and $y_{\text{old}} = y_{\text{new}} + k_0$.

Answer 1

The Correct Option is D

The problem states the roles of the equations in reverse of the standard convention. Let the original coordinates be $(x, y)$ and the new coordinates after shifting the origin be $(X, Y)$.
The origin is shifted to $(-1, 2)$. So, the transformation equations are:
$x = X - 1$ and $y = Y + 2$.
The original equation is $2x^2-xy+y^2-3x+4y-5=0$.
We substitute the transformation equations into this original equation to get the new (transformed) equation in terms of $X$ and $Y$.
$2(X-1)^2 - (X-1)(Y+2) + (Y+2)^2 - 3(X-1) + 4(Y+2) - 5 = 0$.
Expand the terms:
$2(X^2 - 2X + 1) - (XY + 2X - Y - 2) + (Y^2 + 4Y + 4) - 3X + 3 + 4Y + 8 - 5 = 0$.
$2X^2 - 4X + 2 - XY - 2X + Y + 2 + Y^2 + 4Y + 4 - 3X + 3 + 4Y + 8 - 5 = 0$.
Group the terms by powers of $X$ and $Y$:
$2X^2 + Y^2 - XY + (-4-2-3)X + (1+4+4)Y + (2+2+4+3+8-5) = 0$.
$2X^2 - XY + Y^2 - 9X + 9Y + 14 = 0$.
This is the transformed equation. We compare it with the given form $aX^2 + 2hXY + bY^2 + 2gX + 2fY + c = 0$.
$a=2$, $2h=-1 \implies h=-1/2$, $b=1$, $2g=-9 \implies g=-9/2$, $2f=9 \implies f=9/2$, $c=14$.
We need to find the value of $2(f+g+h)$.
$2(f+g+h) = 2(\frac{9}{2} - \frac{9}{2} - \frac{1}{2}) = 2(-\frac{1}{2}) = -1$.
Now we evaluate the expression in the correct option (D): $c-5(a+b)$.
$c-5(a+b) = 14 - 5(2+1) = 14 - 5(3) = 14 - 15 = -1$.
Since both expressions evaluate to -1, the equality holds.