Question

If the tangent at the point P (2,4) 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: This problem deals with the conic sections. A conic section is a curve obtained as the intersection of the surface of a cone with a plane. There are three such types of conic sections which are, the parabola, the hyperbola and the ellipse. This problem is regarding one of those conic sections, which is a parabola. The general form of an equation of a parabola is given by ${y^2} = 4ax$.
The equation of tangent of a parabola $\left( {{x_1},{y_1}} \right)$ is a point on the tangent, is given by:
$ \Rightarrow y{y_1} = 2a\left( {x + {x_1}} \right)$

Complete step-by-step solution:
Given there is a parabola of equation ${y^2} = 8x$, here it is in the standard form of the parabola ${y^2} = 4ax$, hence the given parabola’s vertex is the origin $O(0,0)$.
Here the focal point of the parabola ${y^2} = 8x$ is $\left( {2,0} \right)$, here $a = 2$
Given that there is another parabola of equation ${y^2} = 8x + 5$. The vertex of this parabola does not pass through the origin.
Now given that the tangent to the parabola ${y^2} = 8x$ intersects the parabola ${y^2} = 8x + 5$ at two points Q and R respectively. P (2,4) is a point on the tangent.
We have to find the midpoint of the line QR.
The equation of the tangent to the parabola ${y^2} = 8x$ is given by:
$ \Rightarrow y{y_1} = 2(2)\left( {x + {x_1}} \right)$
As the value of a is 2.
$ \Rightarrow y{y_1} = 4\left( {x + {x_1}} \right)$
Now given P (2,4) is a point on the tangent expressed above, hence substituting the point in the tangent equation as given below:
$ \Rightarrow y(4) = 4\left( {x + 2} \right)$
$ \Rightarrow 4y = 4x + 8$
$ \Rightarrow 4x - 4y + 8 = 0$
$ \Rightarrow x - y + 2 = 0$
Hence the equation of the tangent of the parabola ${y^2} = 8x$ is $x - y + 2 = 0$
The equation of the tangent can be re-written as :
$ \Rightarrow x = y - 2$
As this tangent intersects the parabola ${y^2} = 8x + 5$, hence substituting the value of $x = y - 2$ in this parabola equation:
$ \Rightarrow {y^2} = 8(y - 2) + 5$
$ \Rightarrow {y^2} = 8y - 16 + 5$
\[ \Rightarrow {y^2} - 8y + 11 = 0\]
Solving the quadratic expression to get the values of $y$:
\[ \Rightarrow y = \dfrac{{ - ( - 8) \pm \sqrt {{{\left( 8 \right)}^2} - 4\left( 1 \right)\left( {11} \right)} }}{{2\left( 1 \right)}}\]
\[ \Rightarrow y = \dfrac{{8 \pm \sqrt {20} }}{2}\]
The value of $\sqrt {20} = 4.47$, as given below:
\[ \Rightarrow y = \dfrac{{8 \pm 4.47}}{2}\]
\[ \Rightarrow {y_1} = \dfrac{{8 + 4.47}}{2};{y_2} = \dfrac{{8 - 4.47}}{2}\]
\[ \Rightarrow {y_1} = 6.235;{y_2} = 1.765\]
To get the values of $x$, substitute the obtained values of $y$ in $x = y - 2$ , as given below:
\[ \Rightarrow {x_1} = 6.235 - 2;\]
\[ \Rightarrow {x_2} = 1.765 - 2\]
$\therefore {x_1} = 4.235;$
$\therefore {x_2} = - 0.235$.
Hence the points Q and R are $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ respectively.
Hence Q = $\left( {4.235,6.235} \right)$ , R = $\left( { - 0.235,1.765} \right)$
Now finding the midpoint of QR is given by:
$ \Rightarrow \left( {\left( {\dfrac{{4.235 + ( - 0.235)}}{2}} \right),\left( {\dfrac{{6.235 + 1.765}}{2}} \right)} \right)$
\[ \Rightarrow \left( {\dfrac{4}{2},\dfrac{8}{2}} \right)\]
\[ \Rightarrow \left( {2,4} \right)\]
The midpoint of QR is \[\left( {2,4} \right)\].

Option A is the correct answer.
Note: Please note that while solving this problem we came across various formulas, such as quadratic expression formula and the midpoint formula.
Given a quadratic equation $a{x^2} + bx + c = 0$, then the roots of the equation is given by:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
There is a line joining the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, then the midpoint of the line is given by:
\[ \Rightarrow \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]