6.8 Q-1

Question Statement

Find an equation of each of the following curves with respect to the new parallel axes obtained by shifting the origin to the indicated point.

i. $\quad \mathrm{x}^{2}+16 \mathrm{y}-16=0 \quad 0^{\prime}(0,1)$

ii. $\quad 4 x^{2}+y^{2}+16 x-10 y+37=0 \quad 0^{\prime}(-2,5)$

iii. $\quad 9 x^{2}+4 y^{2}+18 x-16 y-11=0 \quad 0^{\prime}(-1,2)$

iv. $\quad x^{2}-y^{2}+4 x+8 y-11=0 \quad 0^{\prime}(-2,4)$

v. $\quad 9 x^{2}-4 y^{2}+36 x+8 y-4=0 \quad 0^{\prime}(-2,1)$

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Background and Explanation

In this exercise, the goal is to find the equations of conic sections when the origin is shifted to new coordinates. The transformation equations for shifting the origin are:

• $x = x' + h$
• $y = y' + k$

where $(h, k)$ is the new origin. By substituting these transformations into the original equation, the new equation can be derived. This process helps us reframe the equation in terms of the new origin.

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Solution

i. Given Equation:

$\mathrm{x}^{2}+16 \mathrm{y}-16=0 \quad 0^{\prime}(0,1)$

The transformation equations are:

$$x' = x + 0, \quad y' = y + 1$$

Substituting into the original equation:

$$x^{2} + 16(y' - 1) - 16 = 0$$ $$x^{2} + 16y' - 16 - 16 = 0$$ $$x^{2} + 16y' = 0$$

Thus, the transformed equation is:

$$x^{2} + 16y' = 0$$

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ii. Given Equation:

$4 x^{2}+y^{2}+16 x-10 y+37=0 \quad 0^{\prime}(-2,5)$

The transformation equations are:

$$x' = x - 2, \quad y' = y + 5$$

Substituting into the original equation:

$$4(x' + 2)^{2} + (y' - 5)^{2} + 16(x' + 2) - 10(y' - 5) + 37 = 0$$

Expanding and simplifying:

$$4(x'^{2} + 4x' + 4) + (y'^{2} - 10y' + 25) + 16x' + 32 - 10y' + 50 + 37 = 0$$ $$4x'^{2} + 16x' + 16 + y'^{2} - 10y' + 25 + 16x' - 10y' + 45 = 0$$ $$4x'^{2} + y'^{2} - 4 = 0$$

Thus, the transformed equation is:

$$4x'^{2} + y'^{2} - 4 = 0$$

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iii. Given Equation:

$9 x^{2}+4 y^{2}+18 x-16 y-11=0 \quad 0^{\prime}(-1,2)$

The transformation equations are:

$$x' = x - 1, \quad y' = y + 2$$

Substituting into the original equation:

$$9(x' + 1)^{2} + 4(y' - 2)^{2} + 18(x' + 1) - 16(y' - 2) - 11 = 0$$

Expanding and simplifying:

$$9(x'^{2} + 2x' + 1) + 4(y'^{2} - 4y' + 4) + 18x' + 18 - 16y' + 32 - 11 = 0$$ $$9x'^{2} + 18x' + 9 + 4y'^{2} - 16y' + 16 + 18x' + 18 - 16y' + 32 - 11 = 0$$ $$9x'^{2} + 4y'^{2} - 36 = 0$$

Thus, the transformed equation is:

$$9x'^{2} + 4y'^{2} - 36 = 0$$

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iv. Given Equation:

$x^{2}-y^{2}+4 x+8 y-11=0 \quad 0^{\prime}(-2,4)$

The transformation equations are:

$$x' = x - 2, \quad y' = y + 4$$

Substituting into the original equation:

$$(x' + 2)^{2} - (y' - 4)^{2} + 4(x' + 2) + 8(y' - 4) - 11 = 0$$

Expanding and simplifying:

$$(x'^{2} + 4x' + 4) - (y'^{2} - 8y' + 16) + 4x' + 8 + 8y' - 32 - 11 = 0$$ $$x'^{2} - y'^{2} + 1 = 0$$

Thus, the transformed equation is:

$$x'^{2} - y'^{2} + 1 = 0$$

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v. Given Equation:

$9 x^{2}-4 y^{2}+36 x+8 y-4=0 \quad 0^{\prime}(-2,1)$

The transformation equations are:

$$x' = x - 2, \quad y' = y + 1$$

Substituting into the original equation:

$$9(x' + 2)^{2} - 4(y' - 1)^{2} + 36(x' + 2) + 8(y' - 1) - 4 = 0$$

Expanding and simplifying:

$$9(x'^{2} + 4x' + 4) - 4(y'^{2} - 2y' + 1) + 36x' + 72 + 8y' + 8 - 4 = 0$$ $$9x'^{2} - 4y'^{2} - 36 = 0$$

Thus, the transformed equation is:

$$9x'^{2} - 4y'^{2} - 36 = 0$$

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Key Formulas or Methods Used

• Transformation Equations:
  • $x' = x + h$
  • $y' = y + k$
• Substitution Method:
  • Substitute the transformation equations into the given equation to get the new equation with respect to the new axes.

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Summary of Steps

1. Identify the new origin $(h, k)$.
2. Apply the transformation equations $x' = x + h$ and $y' = y + k$.
3. Substitute the transformed coordinates into the given equation.
4. Simplify the equation to get the equation of the curve in the new coordinate system.

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