Question

If the tangent at the point \[P\left( {2,4} \right)\] to the parabola \[{y^2} = 8x\] meets the parabola \[{y^2} = 8x + 5\] at \[Q\] and \[R\], then the midpoint of \[QR\] is
A. \[\left( {2,4} \right)\]
B. \[\left( {4,2} \right)\]
C. \[\left( {7,9} \right)\]
D. None

Answer 1

- Hint: First of all, find the tangent of the parabola \[{y^2} = 8x\]. Then solve the formed tangent and the other parabola to find their points of intersection. And use the midpoint formula to find the required answer.

Complete step-by-step solution -

Given parabola: \[{y^2} = 8x..........................................\left( 1 \right)\]
                              \[{y^2} = 8x + 5..........................................\left( 2 \right)\]
We know that the tangent of the parabola \[{y^2} = 4ax\] at point \[\left( {{x_1},{y_1}} \right)\] is given by \[y{y_1} = 2a\left( {x + {x_1}} \right)\].
So, the tangent of the parabola \[{y^2} = 8x\] at point \[P\left( {2,4} \right)\] is
\[
  4y = 2 \times 2\left( {x + 2} \right) \\
  4y = 4\left( {x + 2} \right) \\
  \therefore y = x + 2.................................................\left( 3 \right) \\
\]
Given the point of intersection of the tangent \[y = x + 2\] and the parabola \[{y^2} = 8x + 5\] are \[Q\] and \[R\].
By solving equation (2) and (3), we get the point of intersection i.e., \[Q\] and \[R\]
\[
   \Rightarrow {\left( {x + 2} \right)^2} = 8x + 5 \\
   \Rightarrow {x^2} + 4x + 4 = 8x + 5 \\
   \Rightarrow {x^2} + 4x - 8x + 4 - 5 = 0 \\
   \Rightarrow {x^2} - 4x - 1 = 0 \\
\]
We know that the roots of the quadratic equation \[a{x^2} + bx + c = 0\] is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
So, the roots of the equation \[{x^2} - 4x - 1 = 0\] is
\[
  x = \dfrac{{4 \pm \sqrt {{4^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}} \\
  x = \dfrac{{4 \pm \sqrt {16 + 4} }}{2} \\
  x = \dfrac{{4 \pm \sqrt {4 \times 5} }}{2} \\
  x = \dfrac{{4 \pm 2\sqrt 5 }}{2} \\
  \therefore x = 2 \pm \sqrt 5 \\
\]
From equation (3), if \[x = 2 - \sqrt 5 \] then \[y = 2 - \sqrt 5 + 2 = 4 - \sqrt 5 \]
From equation (3), if \[x = 2 + \sqrt 5 \] then \[y = 2 + \sqrt 5 + 2 = 4 + \sqrt 5 \]
So, the points of intersection are \[Q\left( {2 - \sqrt 5 ,4 - \sqrt 5 } \right){\text{ and }}R\left( {2 + \sqrt 5 ,4 + \sqrt 5 } \right)\]
Hence the midpoint of \[QR\] is
\[
  \left( {\dfrac{{2 - \sqrt 5 + 2 + \sqrt 5 }}{2},\dfrac{{4 - \sqrt 5 + 4 + \sqrt 5 }}{2}} \right) \\
  \left( {\dfrac{{2 + 2}}{2},\dfrac{{4 + 4}}{2}} \right) \\
  \left( {\dfrac{4}{2},\dfrac{8}{2}} \right) \\
  \left( {2,4} \right) \\
\]
Thus, the correct option is A. \[\left( {2,4} \right)\]

Note: The midpoints of the points \[\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]. The roots of the quadratic equation \[a{x^2} + bx + c = 0\] is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. The transverse axis of both the given parabolas is x-axis.