Question

an ellipse has the equation \\(\frac{(x - 2)^2}{3^2} + \frac{(y + 5)^2}{5^2} = 1\\). if the ellipse is shifted two spaces to the left and five spaces up, what is the new equation?\\(\frac{(x + 2)^2}{3^2} + \frac{(y - 5)^2}{5^2} = 1\\)\\(\frac{x^2}{3^2} + \frac{y^2}{5^2} = 1\\)\\(\frac{(x + 3)^2}{3^2} + \frac{(y + 3)^2}{5^2} = 1\\)\\(\frac{(x - 4)^2}{3^2} + \frac{(y + 10)^2}{5^2} = 1\\)

Explanation:

Step1: Recall the transformation rules for ellipses

For a horizontal shift \( h \) units left, we replace \( x \) with \( x + h \) in the equation of the ellipse. For a vertical shift \( k \) units up, we replace \( y \) with \( y - k \) in the equation of the ellipse. The original equation of the ellipse is \(\frac{(x - 2)^2}{3^2}+\frac{(y + 5)^2}{5^2}=1\). We need to shift it 2 units left and 5 units up.

Step2: Apply the horizontal shift

Shifting 2 units left means \( h = 2 \), so we replace \( x \) with \( x+2 \) in the \( x \)-term. The \( x \)-term was \((x - 2)\), after replacing \( x \) with \( x + 2\), we get \(( (x + 2)- 2)=(x)\)? Wait, no. Wait, the original \( x \)-coordinate of the center is \( 2 \) (from \((x - 2)\)). Shifting left by 2 units: new \( x \)-coordinate of center is \( 2- 2=0 \)? Wait, no, the transformation for horizontal shift: if we shift the graph of \( y = f(x) \) left by \( a \) units, we get \( y = f(x + a) \). So for the ellipse equation \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), shifting left by \( a \) units: \( h\) becomes \( h - a \), so the \( x \)-term becomes \((x-(h - a))=(x - h + a)\). Wait, original center is \((h,k)=(2,- 5)\). Shifting left by 2 units: new \( h'=h - 2=2 - 2 = 0\)? No, wait, shifting left by 2 units: the \( x \)-coordinate of the center moves from \( x = 2 \) to \( x=2 - 2=0 \)? Wait, no, left shift: if you have a point \((x,y)\) on the original ellipse, after shifting left by 2 units, the new point is \((x - 2,y)\)? Wait, I think I messed up. Let's correct:

The standard form of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), where \((h,k)\) is the center.

- Shifting the ellipse \( c \) units to the left: the new center is \((h - c,k)\), so the \( x \)-term becomes \((x-(h - c))=(x - h + c)\)
- Shifting the ellipse \( d \) units up: the new center is \((h,k + d)\), so the \( y \)-term becomes \((y-(k + d))=(y - k - d)\)

Original center: \( h = 2\), \( k=-5 \)

Shifting 2 units left: \( c = 2\), new \( h'=h - c=2 - 2 = 0\)? No, wait, shifting left by 2 units: the \( x \)-coordinate of the center decreases by 2. So original \( h = 2 \), new \( h'=2-2 = 0\)? Wait, no, if we shift the graph left by 2 units, the transformation on the \( x \)-variable is \( x\to x + 2 \) (because \( f(x)\to f(x + 2) \) for left shift). So for the \( x \)-term \((x - h)\), replacing \( x \) with \( x + 2 \), we get \(( (x + 2)-h)\). Original \( h = 2 \), so \((x + 2 - 2)=x\). Wait, that can't be. Wait, let's take an example: original ellipse center at (2, -5). Shift left by 2 units: new center is (2 - 2, -5)=(0, -5). Shift up by 5 units: new center is (0, -5 + 5)=(0, 0). Wait, no, the problem says shift two spaces to the left and five spaces up.

Wait, original equation: \(\frac{(x - 2)^2}{3^2}+\frac{(y + 5)^2}{5^2}=1\). Let's rewrite \( y + 5\) as \( y-(-5) \), so center is (2, -5).

Shifting left by 2 units: the \( x \)-coordinate of the center: \( 2-2 = 0\)

Shifting up by 5 units: the \( y \)-coordinate of the center: \(-5 + 5=0\)

Wait, no, shifting up by 5 units: the \( y \)-coordinate increases by 5, so from \( y=-5 \) to \( y=-5 + 5 = 0\)

Now, the new center is (0, 0)? Wait, but let's check the options. Wait, the first option after the original is \(\frac{(x + 2)^2}{3^2}+\frac{(y - 5)^2}{5^2}=1\)

Wait, maybe my transformation was wrong. Let's use the function transformation:

If we have a graph \( \frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1 \)

- Shift left by \( m \) units: replace \( x \) with \( x + m \) in the \( x \)-term. So \((x - h)\) becomes \(( (x + m)-h)=(x-(h - m))\)
- Shift up by \( n \) units: replace \( y \) with \( y - n \) in the \( y \)-term. So \((y - k)\) becomes \(( (y - n)-k)=(y-(k + n))\)

Original equation: \( h = 2\), \( k=-5 \), \( m = 2\) (left shift), \( n = 5\) (up shift)

So for the \( x \)-term: \((x - h)\) becomes \((x-(h - m))=(x-(2 - 2))=x\)? No, wait, original \( x \)-term: \((x - 2)\). After shifting left by 2 units, we have \( (x + 2 - 2)=x \)? No, that's not right. Wait, let's take a point on the original ellipse. Let's take the center (2, -5). After shifting left by 2 units and up by 5 units, the center moves to (2 - 2, -5 + 5)=(0, 0). So the new equation should be \(\frac{(x - 0)^2}{3^2}+\frac{(y - 0)^2}{5^2}=1\)? But that's \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\), which is one of the options? Wait, no, the first option after the original is \(\frac{(x + 2)^2}{3^2}+\frac{(y - 5)^2}{5^2}=1\)

Wait, maybe I made a mistake in the direction of the shift. Let's re-express the transformation:

- Shifting a graph \( y = f(x) \) left by \( a \) units: \( y = f(x + a) \)
- Shifting a graph \( y = f(x) \) up by \( b \) units: \( y = f(x)+b \)

So for the ellipse equation, which is a function of \( x \) and \( y \), the equation is \(\frac{(x - 2)^2}{9}+\frac{(y + 5)^2}{25}=1\)

To shift left by 2 units: replace \( x \) with \( x + 2 \) in the \( x \)-term. So \( (x - 2) \) becomes \( ( (x + 2)- 2)=x \)? No, that's not. Wait, \( (x - 2) \) when \( x \) is replaced by \( x + 2 \) is \( ( (x + 2)- 2)=x \). So the \( x \)-term becomes \( x \), so \(\frac{x^2}{3^2}\). For the \( y \)-term: \( (y + 5) \) is \( (y - (-5)) \). Shifting up by 5 units: replace \( y \) with \( y - 5 \) (because shifting up by 5 units: \( y \) becomes \( y - 5 \) to get the same \( y \)-value as before). So \( (y + 5) \) becomes \( ( (y - 5)+ 5)=y \). So the \( y \)-term becomes \( y \), so \(\frac{y^2}{5^2}\). So the new equation is \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\), which is the option \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\)

Wait, but let's check the options again. The options are:

1. \(\frac{(x + 2)^2}{3^2}+\frac{(y - 5)^2}{5^2}=1\)

2. \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\)

3. \(\frac{(x + 3)^2}{3^2}+\frac{(y + 3)^2}{5^2}=1\)

4. \(\frac{(x - 4)^2}{3^2}+\frac{(y + 10)^2}{5^2}=1\)

Ah, so the second option is \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\), which is \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\)

Wait, but let's re-examine the shift. The original center is (2, -5). Shifting left by 2: subtract 2 from the x-coordinate: 2 - 2 = 0. Shifting up by 5: add 5 to the y-coordinate: -5 + 5 = 0. So the new center is (0, 0), so the equation is \(\frac{(x - 0)^2}{3^2}+\frac{(y - 0)^2}{5^2}=1\), which is \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\), which is the second option.

Wait, but maybe I messed up the sign in the transformation. Let's use the standard transformation for conic sections:

The general equation of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), where \((h,k)\) is the center.

- To shift the ellipse \( c \) units to the left: the new center is \((h - c, k)\), so the equation becomes \(\frac{(x-(h - c))^2}{a^2}+\frac{(y - k)^2}{b^2}=1\)

- To shift the ellipse \( d \) units up: the new center is \((h, k + d)\), so the equation becomes \(\frac{(x - h)^2}{a^2}+\frac{(y-(k + d))^2}{b^2}=1\)

Combining both shifts: new center is \((h - c, k + d)\), so the equation is \(\frac{(x-(h - c))^2}{a^2}+\frac{(y-(k + d))^2}{b^2}=1\)

Given \( h = 2\), \( k=-5\), \( c = 2\) (left shift), \( d = 5\) (up shift)

So \( h - c = 2 - 2 = 0\), \( k + d=-5 + 5 = 0\)

Thus, the new equation is \(\frac{(x - 0)^2}{3^2}+\frac{(y - 0)^2}{5^2}=1\), which is \(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\)

Answer:

\(\frac{x^2}{3^2}+\frac{y^2}{5^2}=1\) (the second option: \(\boldsymbol{\frac{x^2}{3^2}+\frac{y^2}{5^2}=1}\))