SOLUTION: Find the slope of the tangent line: (a)To the ellipse x^2+y^2/2=1 at the point (1/√2,1) (b)To the hyperbola x^2−y^2=1 at the point (√3,√2)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the slope of the tangent line: (a)To the ellipse x^2+y^2/2=1 at the point (1/√2,1) (b)To the hyperbola x^2−y^2=1 at the point (√3,√2)       Log On


   

Question 825587: Find the slope of the tangent line:
(a)To the ellipse x^2+y^2/2=1 at the point (1/√2,1)
(b)To the hyperbola x^2−y^2=1 at the point (√3,√2)

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Find the slope of the tangent line:
 (a)To the ellipse x^2+y^2/2=1 at the point (1/√2,1) 
 (b)To the hyperbola x^2−y^2=1 at the point (√3,√2) 

The slope of the tangent line at a point IS the derivative
at that point.

x%5E2%2By%5E2%2F2%22%22=%22%221

Clear of fractions by multiplying through by 2

2x%5E2%2By%5E2%22%22=%22%222

Take the derivative term-by-term implicitly:

4x%2Aexpr%28%28dx%29%2F%28dx%29%29%2B2y%2Aexpr%28%28dy%29%2F%28dx%29%29%22%22=%22%22%220%22

expr%28%28dx%29%2F%28dx%29%29 is just 1 so

4x%2B2y%2Aexpr%28%28dy%29%2F%28dx%29%29%22%22=%22%22%220%22

Subtract 4x from both sides:

2y%2Aexpr%28%28dy%29%2F%28dx%29%29%22%22=%22%22-4x

Divide both sides by 2y

%28dy%29%2F%28dx%29%22%22=%22%22-4x%2F%282y%29

%28dy%29%2F%28dx%29%22%22=%22%22-%282x%29%2Fy

So we substitute the point (1%2Fsqrt%282%29,1)

%28dy%29%2F%28dx%29@(1%2Fsqrt%282%29,1) %22%22=%22%22%28-2%281%2Fsqrt%282%29%29%29%2F1%22%22=%22%22-2%2Fsqrt%282%29%22%22=%22%22expr%28-2%2Fsqrt%282%29%29expr%28sqrt%282%29%2Fsqrt%282%29%29%22%22=%22%22-2%28sqrt%282%29%29%2F2%22%22=%22%22-sqrt%282%29

The other one is done the same way.

Edwin