Chord and tangent 

In this page 'Chord and tangent' we are going to derive the equation of tangent and equation of chord to the given ellipse.                    

Derivation of the equation chord and tangent:  

Let x²/a² + y²/b² =1 is the equation of the ellipse. Let

 P(x₁,y₁) be the given point and Q(x₂,y₂) be a point on the ellipse close to P.

                   The equation of PQ is

                      y- y₁      = [(y₁-y₂)/(x₁-x₂)](x-x₁) ----------- I   

                    Since P(x₁,y₁) and Q(x₂,y₂) lie on the ellipse.

                     x₁²/a² + y₁²/b² =1 and x₂²/a² + y₂²/b² =1.

                     By subtracting,

                     (x₁²-x₂²)/a²  +  ( y₁²-y₂²)/b²     =  0

                     (x₁+x₂)(x₁-x₂)/a²  +  (y₁+y₂)(y₁-y₂)/b²  =  0

                     (y₁+y₂)(y₁-y₂)/b²    =    -  (x₁+x₂)(x₁-x₂)/a² 

                     (y₁-y₂)/(x₁-x₂)        =    -   b²/a²  .  (x₁+x₂)/(y₁+y₂)----II

                     Substituting II in I  

                              y-y₁              =    - (b²/a²)[(x₁+x₂)/(y₁+y₂)] . (x-x₁)

                    [x(x₁+x₂)]/a²  + [y(y₁+y₂)]/b²  =  [(x₁ . x₂)/a²] +        

                                                                            [(y₁.y₂)/b²] + 1


                   This is the equation of the chord PQ. 

                   When Q approaches P, the chord becomes  tangent at P. 

                   y₂ becomes y₁ and x₂ becomes x₁.

                   So substituting  y₂= y₁ and   x₂= x₁, we get the equation of the tangent at P.

                     y- y₁  =    -   b²/a² .  x₁/ y₁  .  (x- x₁)

                     (xx₁/ a²)  +  (yy₁/b²) =  x₁²/a²  + y₁²/b²  = 1. 

                  Thus the equation of tangent at P(x₁,y₁) is  

                              (xx₁/ a²)  +  (yy₁/b²)    = 1                     

Condition for a line y= mx+c to be a tangent to the ellipse.

Let us see the condition for the line y=mx+c to be a tangent to the ellipse   x²/a² + y²/b² =1.

From the two equations let us eliminate y, we get the equation as

         [x²(1/a² + m/b²)]+[2mcx/b²]+[(c²/b²)-1] =0

We got a quadratic equation here. If the straight line y=mx+c is a tangent then the discriminant of the above equation is 0.

         (2mc/b²)² -4[(1/a² + m/b²)[(c²/b²)-1]  = 0

         4m²c²/b⁴    -4[c²/a²b² - 1/a² + m²c²/b⁴  -m²/b²]   = 0

                           c²  =    a²m² +  b²

                           c   =     ±√( a²m² +  b²)

      This is the required condition.

      The point of contact is [ -a²m/√( a²m² +  b²), b²/√( a²m² +  b²)].

      The equation of tangents will be y = mx  ±√( a²m² +  b²)

Parents and teachers can guide the students to follow the steps of derivation of  equations in this page 'Chord and tangent'  and examples(finding the equation of tangent and chord) in the next page. Students can follow the examples step by step and try to derive the equations on their own. If you have any doubt you can contact us through mail, we will help you to clear your doubts. 



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