- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
Implicit differentiation and inverse trigonometric functions practice question 24
The ellipse below has exactly two tangent lines parallel to the line . Find an equation of the one that touches the ellipse at a point with .
What slope must a tangent parallel to have? Where on the ellipse does the tangent have that slope?
Parallel to means slope . Find implicitly and set it equal to .
That condition is a relation between and . Substitute it into the ellipse to find the points of tangency, and keep the one with .
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
Parallel lines share a slope. Setting the implicit derivative equal to −1 gives a line through the origin on which the points of tangency lie; intersect it with the ellipse, keep the point with x > 0, and write the tangent line there.
Working
Implicit differentiation; the chain rule turns 3y² into 6y dy/dx.
Slope −1, the slope of y = −x.
Substitute x = 3y into the ellipse.
Both points lie on the ellipse: 9 + 3 = 12.
Point-slope form with slope −1.
Answer
The answer is y equals negative x plus 4.
Check
Substituting y = 4 − x into x² + 3y² = 12 leaves 4(x − 3)² = 0, a double root, so the line touches the ellipse once, at (3, 1).
Where this goes wrong
Losing the minus sign in dy/dx = −x/(3y), which leads to x = −3y and the point (3, −1), where the slope is +1. Set the derivative equal to −1 exactly, then keep the point with x > 0.
Leans on
Equation of a tangent line, from The definition of the derivative
Try a similar question
Find an equation of the tangent line to the ellipse below at the point .Saved in this browser only.