Find the condition so that the line p x + q y = r intersects the ellipse x 2 a 2 + y 2 b 2 = 1 in points whose eccentric angles differ by π 4 Though I know how to solve it using parametric coordinates, I was wondering if there's an another approach which is less time consuming.

Question

Find the condition so that the line   p  x  +  q  y  =  r intersects the ellipse             x      2              a      2        +            y      2              b      2        =  1 in points whose eccentric angles differ by       π    4  Though I know how to solve it using parametric coordinates, I was wondering if there&#039;s an another approach which is less time consuming.

bosco84jsrztu · Accepted Answer

Step 1Let       P    1    =  (  a  cos  ⁡  t  ,  b  sin  ⁡  t  ) and       P    2    =  (  a  cos  ⁡  (  t  +  ϕ  )  ,  b  sin  ⁡  (  t  +  ϕ  )  ) where   ϕ  =            π      4       , then we need to satisfy two conditions:1)       P    1   is on the line   p  x  +  q  y  =  r , therefore,  p  a  cos  ⁡  t  +  q  b  sin  ⁡  t  =  r    (  1  )2)       P    2   is on the line as well, hence,  p  a  cos  ⁡  (  t  +  ϕ  )  +  q  b  sin  ⁡  (  t  +  ϕ  )  =  r    (  2  )This last equation becomes  cos  ⁡  t  (  p  a  cos  ⁡  ϕ  +  q  b  sin  ⁡  ϕ  )  +  sin  ⁡  t  (  −  p  a  sin  ⁡  ϕ  +  q  b  cos  ⁡  ϕ  )  =  r    (  3  )And after substituting   cos  ⁡  ϕ  =  sin  ⁡  ϕ  =            1              2             , it becomes,  cos  ⁡  t  (  p  a  +  q  b  )  +  sin  ⁡  t  (  q  b  −  p  a  )  =      2    r    (  4  )Now solving (1) and (4) for cos t and sin t, we find that  cos  ⁡  t  =  −  r                    (        q        b        −        p        a        )        −                  2                q        b                              p          2                          a          2                +                  q          2                          b          2                      sin  ⁡  t  =  −  r                              2                p        a        −        (        p        a        +        q        b        )                              p          2                          a          2                +                  q          2                          b          2                    Since       cos    2    ⁡  t  +      sin    2    ⁡  t  =  1 , then the condition is                    (                  p          2                          a          2                +                  q          2                          b          2                          )          2                            r        2              =  (  q  b  −  p  a      )    2    +  (  p  a  +  q  b      )    2    +  2  (      q    2        b    2    +      p    2        a    2    )  −  2      2    (  q  b  (  q  b  −  p  a  )  +  p  a  (  p  a  +  q  b  )  )Simplifying the right hand side,                    (                  p          2                          a          2                +                  q          2                          b          2                          )          2                            r        2              =  4  (      p    2        a    2    +      q    2        b    2    )  −  2      2    (      q    2        b    2    +      p    2        a    2    )Dividing through by   (      p    2        a    2    +      q    2        b    2    ) , yields,                              p          2                          a          2                +                  q          2                          b          2                            r        2              =  4  −  2      2  And this is the condition on the line.

Find the condition so that the line p x + q y = r intersects the ellipse

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