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Math  /  Calculus

Question1. Find the equations of the tangent line to the ellipse, (x+2)24+(y−3)29\frac{(x+2)^{2}}{4}+\frac{(y-3)^{2}}{9} when x=−3x=-3.

Studdy Solution

STEP 1

1. The ellipse is given by the equation (x+2)24+(y−3)29=1\frac{(x+2)^2}{4} + \frac{(y-3)^2}{9} = 1.
2. We need to find the equation of the tangent line at a specific xx-coordinate, x=−3x = -3.
3. The tangent line can be found using the derivative of the ellipse equation.

STEP 2

1. Find the point on the ellipse where x=−3x = -3.
2. Differentiate the ellipse equation implicitly to find the slope of the tangent line.
3. Use the point-slope form to write the equation of the tangent line.

STEP 3

Substitute x=−3x = -3 into the ellipse equation to find the corresponding yy-coordinate(s):
((−3)+2)24+(y−3)29=1\frac{((-3)+2)^2}{4} + \frac{(y-3)^2}{9} = 1
Simplify and solve for yy:
(−1)24+(y−3)29=1\frac{(-1)^2}{4} + \frac{(y-3)^2}{9} = 1 14+(y−3)29=1\frac{1}{4} + \frac{(y-3)^2}{9} = 1 (y−3)29=1−14\frac{(y-3)^2}{9} = 1 - \frac{1}{4} (y−3)29=34\frac{(y-3)^2}{9} = \frac{3}{4} (y−3)2=274(y-3)^2 = \frac{27}{4} y−3=±272y-3 = \pm \frac{\sqrt{27}}{2} y=3±332y = 3 \pm \frac{3\sqrt{3}}{2}
The points on the ellipse are (−3,3+332)(-3, 3 + \frac{3\sqrt{3}}{2}) and (−3,3−332)(-3, 3 - \frac{3\sqrt{3}}{2}).

STEP 4

Differentiate the ellipse equation implicitly with respect to xx:
ddx((x+2)24+(y−3)29)=ddx(1)\frac{d}{dx}\left(\frac{(x+2)^2}{4} + \frac{(y-3)^2}{9}\right) = \frac{d}{dx}(1)
Apply the chain rule:
14⋅2(x+2)+19⋅2(y−3)⋅dydx=0\frac{1}{4} \cdot 2(x+2) + \frac{1}{9} \cdot 2(y-3) \cdot \frac{dy}{dx} = 0
Simplify:
x+22+2(y−3)9⋅dydx=0\frac{x+2}{2} + \frac{2(y-3)}{9} \cdot \frac{dy}{dx} = 0
Solve for dydx\frac{dy}{dx}:
2(y−3)9⋅dydx=−x+22\frac{2(y-3)}{9} \cdot \frac{dy}{dx} = -\frac{x+2}{2} dydx=−9(x+2)4(y−3)\frac{dy}{dx} = -\frac{9(x+2)}{4(y-3)}

STEP 5

Calculate the slope of the tangent line at each point:
1. For the point (−3,3+332)(-3, 3 + \frac{3\sqrt{3}}{2}):
dydx=−9((−3)+2)4(3+332−3)\frac{dy}{dx} = -\frac{9((-3)+2)}{4(3 + \frac{3\sqrt{3}}{2} - 3)} =963= \frac{9}{6\sqrt{3}} =323= \frac{3}{2\sqrt{3}}
2. For the point (−3,3−332)(-3, 3 - \frac{3\sqrt{3}}{2}):
dydx=−9((−3)+2)4(3−332−3)\frac{dy}{dx} = -\frac{9((-3)+2)}{4(3 - \frac{3\sqrt{3}}{2} - 3)} =−963= -\frac{9}{6\sqrt{3}} =−323= -\frac{3}{2\sqrt{3}}
Now, use the point-slope form to write the equation of the tangent line for each point:
1. For (−3,3+332)(-3, 3 + \frac{3\sqrt{3}}{2}):
y−(3+332)=323(x+3)y - \left(3 + \frac{3\sqrt{3}}{2}\right) = \frac{3}{2\sqrt{3}}(x + 3)
2. For (−3,3−332)(-3, 3 - \frac{3\sqrt{3}}{2}):
y−(3−332)=−323(x+3)y - \left(3 - \frac{3\sqrt{3}}{2}\right) = -\frac{3}{2\sqrt{3}}(x + 3)
The equations of the tangent lines are:
1. y−(3+332)=323(x+3)y - \left(3 + \frac{3\sqrt{3}}{2}\right) = \frac{3}{2\sqrt{3}}(x + 3)
2. y−(3−332)=−323(x+3)y - \left(3 - \frac{3\sqrt{3}}{2}\right) = -\frac{3}{2\sqrt{3}}(x + 3)

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