- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Implicit differentiation and inverse trigonometric functions practice question 6
Find an equation of the tangent line to the ellipse below at the point .
A tangent line needs a point and a slope. The point is given; how do you get the slope from an equation that mixes and ?
Differentiate both sides with respect to : the chain rule turns into . Solve for , substitute the point, then write the line through with that slope.
An equation of a line. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §4.5CLP-1 §2.11Key results
Worked solution
Strategy
Differentiate implicitly to get the slope at (2, 1), then write the line through the point with that slope.
Working
The chain rule on 4y²; the constant 8 differentiates to 0.
Substitute both coordinates.
Point-slope form, then solved for y.
Answer
The answer is y equals negative one half x plus 2.
Check
Substituting y = 2 − x/2 into x² + 4y² = 8 leaves 2(x − 2)² = 0, a double root, so the line touches the ellipse only at (2, 1).
Where this goes wrong
Turning the ratio upside down, dy/dx = −8y/(2x), or losing the 4 on 4y². Either gives slope −2. Divide by the whole coefficient of dy/dx, 8y.
Leans on
Equation of a tangent line, from The definition of the derivative
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