To solve the equation $x^2 + 16y^2 - 6x - 96y + 144 = 7$, we want to rewrite it in a standard form, which usually involves completing the square for both $x$ and $y$.

Let's go step-by-step:

Step 1: Rearrange the equation

Move the constant term to the right side:

$x^2 + 16y^2 - 6x - 96y + 144 = 7$

$x^2 - 6x + 16y^2 - 96y = 7 - 144$

$x^2 - 6x + 16y^2 - 96y = -137$

Step 2: Complete the square for $x$ and $y$

For $x$:

Take the terms involving $x$: $x^2 - 6x$

To complete the square:

$x^2 - 6x = (x - 3)^2 - 9$

For $y$:

Take the terms involving $y$: $16y^2 - 96y$

Factor out the 16:

$16(y^2 - 6y)$

Now, complete the square inside the parentheses:

$y^2 - 6y = (y - 3)^2 - 9$

Therefore,

$16(y^2 - 6y) = 16((y - 3)^2 - 9) = 16(y - 3)^2 - 144$

Step 3: Rewrite the equation

Substitute the completed square forms back into the equation:

$(x - 3)^2 - 9 + 16(y - 3)^2 - 144 = -137$

Combine the constants:

$(x - 3)^2 + 16(y - 3)^2 - 153 = -137$

$(x - 3)^2 + 16(y - 3)^2 = 16$

Step 4: Write in standard form

$\frac{(x - 3)^2}{16} + \frac{(y - 3)^2}{1} = 1$

Conclusion

This is the standard form of an ellipse with the center at $(3, 3)$, a semi-major axis of length 4 (along the x-direction), and a semi-minor axis of length 1 (along the y-direction).

Would you like further details or have any questions?

Related Questions:

1. How to find the foci of the ellipse from the standard form?
2. What is the area of the ellipse given its semi-major and semi-minor axes?
3. How do you graph an ellipse given its standard form equation?
4. What transformations are needed to shift the ellipse back to the origin?
5. How to find the eccentricity of the ellipse?

Tip: When completing the square, always remember to balance the equation by adding or subtracting the same value on both sides!