Find the equation of an ellipse if its center is S(2, 1) and the edges of a triangle PQR are tangent lines to this ellipse. P(0, 0), Q(5, 0), R(0, 4).My attempt: Let take a point on the line PQ. For example (m,0). Then we have an equation of a tangent line for this point: ( a 11 m + a 1 ) x + ( a 12 m + a 2 ) y + ( a 1 m + a ) = 0, where a 11 etc are coefficients of our ellipse: a 11 x 2 + 2 a 12 x y + a 22 y 2 + 2 a 1 x + 2 a 2 y + a = 0. Now if PQ: y = 0, then ( a 11 m + a 1 ) = 0 , a 12 m + a 2 = 1, a 1 m + a = 0 .I've tried this method for other 2 lines PR and RQ and I got 11 equations (including equations of a center)! Is there a better solution to this problem?

Question

Find the equation of an ellipse if its center is S(2, 1) and the edges of a triangle PQR are tangent lines to this ellipse. P(0, 0), Q(5, 0), R(0, 4).My attempt: Let take a point on the line PQ. For example (m,0). Then we have an equation of a tangent line for this point:   (      a          11        m  +      a    1    )  x  +  (      a          12        m  +      a    2    )  y  +  (      a    1    m  +  a  )  =  0, where       a          11       etc are coefficients of our ellipse:       a          11            x    2    +  2      a          12        x  y  +      a          22            y    2    +  2      a    1    x  +  2      a    2    y  +  a  =  0. Now if PQ: y = 0, then   (      a          11        m  +      a    1    )  =  0 ,       a          12        m  +      a    2    =  1,       a    1    m  +  a  =  0 .I&#039;ve tried this method for other 2 lines PR and RQ and I got 11 equations (including equations of a center)! Is there a better solution to this problem?

Braylon Perez · Accepted Answer

Step 1Translate the center to the origin.. the triangle translates toThe translated points become   (  −  2  ,  −  1  )  ,  (  3  ,  −  1  )  ,  (  −  2  ,  3  )I considered this on the projective plane, I don&#039;t know if that makes it any easier, but that is what I did... if you don&#039;t know projective geometry, just set   z  =  1 in everything that follows.      [                            x                          y                          z                      ]        [                            A                          B                          0                                      B                          C                          0                                      0                          0                          D                      ]        [                            x                                      y                                      z                      ]    =  0Describes a cone, which when it intersects the plane   z  =  1 forms the ellipsei.e.   A      x    2    +  2  B  x  y  +  C      y    2    +  D  =  0The planes   x  +  2  z  =  0  ,  y  +  z  =  0 and   4  x  +  5  y  −  7  z  =  0 are tangent to the cone.Consider the triplet   (  B  ,  −  A  ,  1  )      [                            A                          B                          0                                      B                          C                          0                                      0                          0                          D                      ]        [                            B                                      −          A                                      1                      ]    =      [                            0                                                  B            2                    −          A          C                                      D                      ]  If   A  =  1 this point lies in the plane   y  +  z  =  0.. And if   D  =      B    2    −  A  C      [                            B                          −          1                          1                      ]        [                            0                                                  B            2                    −          A          C                                                  B            2                    −          A          C                      ]    =  0The point is on our cone.Similarly   (  −      C    2    ,      B    2    ,  1  )      [                            A                          B                          0                                      B                          C                          0                                      0                          0                                      B            2                    −          A          C                      ]        [                            −                      C            2                                                            B            2                                                1                      ]    =      [                                        1            2                    (                      B            2                    −          A          C          )                                      0                                                  B            2                    −          A          C                      ]  If   C  =  4 the point is on the plane   x  +  2  z  =  0 and on the cone. Our matrix thus far.      [                            1                          B                          0                                      B                          4                          0                                      0                          0                                      B            2                    −          4                      ]  We just need an equation for the 3rd point of tangency.  (            4      C      −      5      B        7    ,            5      A      −      4      B        7    ,  1  )If it is in the plane   4  x  +  5  y  −  7  z  =  0  4            4      C      −      5      B        7    +  5            5      A      −      4      B        7    −  7  =  0    4            16      −      5      B        7    +  5            5      −      4      B        7    −  7  =  0    64  −  40  B  +  25  −  49  =  0    40  −  40  B  =  0    B  =  1And finally we need to check to see if this point is on the cone      [                            1                          1                          0                                      1                          4                          0                                      0                          0                          −          3                      ]        [                                        11            7                                                            1            7                                                1                      ]    =      [                                        12            7                                                            15            7                                                −          3                      ]        [                                        11            7                                                1            7                                    1                      ]    =      [                                        12            7                                                            15            7                                                −          3                      ]    =            132      +      15        49    −  3  =  0      x    2    +  2  x  y  +  4      y    2    −  3  =  0And finally translate back to the original coordinates  (  x  −  2      )    2    +  2  (  x  −  2  )  (  y  −  1  )  +  4  (  y  −  1      )    2    −  3  =  0

preityk7t · Accepted Answer

Step 1The triangle vertices are       P    1    (  0  ,  0  )  ,      Q    1    (  5  ,  0  )  ,      R    1    (  0  ,  4  )The equation of the ellipse in matrix-vector form is  (  r  −  C      )    T    Q  (  r  −  C  )  =  1where   C  =  (  2  ,  1  ) is the center , and Q is a   2  ×  2 symmetric matrix.Drawing the triangle, we realize that the required ellipse is tangent to the x axis, the y axis, and the line   y  =  5  −      5    4    x .Starting with the x axis, we know that the gradient will be pointing in the -j direction.Now the gradient   g  =  2  Q  (  r  −  C  ). If       r    1   is the tangent point to the ellipse on the x axis then we must have  Q  (      r    1    −  C  )  =  α  (  −  j  )So that   (      r    1    −  C  )  =  −  α      Q          −      1        jPluggin this into the ellipse equation gives us   α  =            1                        j          T                          Q                      −            1                          j            Therefore,       r    1    −  C  =  −                              Q                      −            1                          j                              j          T                          Q                      −            1                          j            Note that       r    1   is a point on the x axis, so its y coordinate is zero, i.e.       j    T        r    1    =  0. Hence      j    T    (      r    1    −  C  )  =  −      j    T    C  =  −      C    y    =  −            j      T              Q              −        1              j  Hence,       C    y    =  1  =            j      T              Q              −        1              j    =      Q          22              −      1      Similar to the above reasoning, we find that       C    x    =            i      T              Q              −        1              i    =      Q          11              −      1      Hence, so far,       Q          −      1        =      [                            4                                              Q                          12                                      −              1                                                                        Q                          12                                      −              1                                                        1                      ]  To find the last unknown, we use the the third condition, which is tangency with the line   y  =  5  −      5    4    xThe vector   Q  R  =  (  −  5  ,  4  ) so the unit normal vector to QR is   n  =            1              41              (  4  ,  5  )Similar to the first tangency analysis with the x axis, we have      r    3    −  C  =  γ      Q          −      1        nPlugging this into the ellipse equation results in   γ  =            1                        n          T                          Q                      −            1                          n            Therefore,       n    T    (      r    3    −  C  )  =      n    T    (      R    1    −  C  )  =      n    T    (      Q    1    −  C  )  =            n      T              Q              −        1              n  Now       n    T    (      R    1    −  C  )  =      n    T    (  (  0  ,  4  )  −  (  2  ,  1  )  )  =            1              41              (  (  4  ,  5  )  ⋅  (  −  2  ,  3  )  =            7              41            And,       n    T        Q          −      1        n  =            1      41        (  4  (  4      )    2    +  1  (  5      )    2    +  2      Q          12              −      1        (  4  )  (  5  )  )Hence   40      Q          12              −      1        =  −  40 From which       Q          12              −      1        =  −  1Thus,       Q          −      1        =      [                            4                                  −          1                                      −          1                                  1                      ]  Inverting, we get   Q  =            1      3            [                            1                                  1                                      1                                  4                      ]  Plugging Q in the ellipse equation and expanding gives  (  x  −  2      )    2    +  2  (  x  −  2  )  (  y  −  1  )  +  4  (  y  −  1      )    2    =  3

Find the equation of an ellipse if its center is S(2, 1) and the edges of a triangle PQR are tangent

Answered question

Answer & Explanation

New Questions in High school geometry