More Calculus Answers
Question Library
Ask a question about Calculus
Volunteer
Experts of the Month
Expert Login
Awards
About Us
Tell friends
Link to Us
Disclaimer
|
| |
|
|
| |
| | | |
About Alon Mandes
Expertise
Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems.
Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
Experience
1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
Organizations
Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.
Education/Credentials
M.A in Mathematics & Bs.c in Electronics.
| | |
| |
You are here: Experts > Teens > Homework/Study Tips > Calculus > Lines tangent to ellipse
Calculus - Lines tangent to ellipse
Expert: Alon Mandes - 11/5/2009
Question
Find the equations of both tangent lines to the ellipse ((x^2)/4)+((y^2)/9)=1 that pass through the point (4,0).
Answer
Let's suppose that the point where the tangent line touch the ellipse is (xo,yo). We can find
a relation between xo %26 yo, by substituting (xo,yo) in the equation of the ellipse :
(xoČ/4)+(yoČ/9)=1 --> (yoČ/9)=1-(xoČ/4) --> yoČ=9[1-(xoČ/4)] --> yo=±3√[1-(xoČ/4)].
Let's call this Eq#1 .
The slope of the tangent line is equal to the derivative of the curve at point(xo,yo). Let's
derive : (2xo/4)+(2yo/9)y'=0 --> y'=-9xo/4yo=m . Let's call it Eq#2 . By substituting Eq#1 we gain : m=-9xo/{4*±3√[1-(xoČ/4)]} --> m=±xo/{4√[1-(xoČ/4)]} . Let's call this Eq#3.
The equation of the tangent line is : Y=mX+n . We can find a relation between m %26 n , by
substituting the point(4,0) in the tangent line equation : 0=4m+n . Let's call this Eq#4 .
The tangent line passes also through the point (xo,yo), therefore : yo=mxo+n. Let's call
this equation #5 . So, we have 5 equations with 4 variables. Let's list them :
#1 : yo=±3√[1-(xoČ/4)]
#2 : m=-9xo/4yo --> yo=-9xo/m
#3 : m=±xo/{4√[1-(xoČ/4)]}
#4 : 0=4m+n --> n=-4m
#5 : yo=mxo+n
I will leave it to from here to continue ..
Alon.
Add to this Answer Ask a Question
|
|
