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.