Analytic Geometry - Parabola

If the line through $Q(1, 1)$ has a slope of $m$ , then its equation is $y - 1 = m(x - 1)$ , and it intersects the parabola $y^2 = 4x$ at $D(\frac{2 + 2\sqrt{m^2 - m + 1} + m^2 - m}{m^2}, \frac{2 + 2\sqrt{m^2 - m + 1}}{m})$ and $E(\frac{2 - 2\sqrt{m^2 - m + 1} + m^2 - m}{m^2}, \frac{2 - 2\sqrt{m^2 - m + 1}}{m})$ .

The line through $R(1, 2)$ and $D$ has an equation of $y - 2 = \frac{D_y - 2}{D_x - 1}(x - 1)$ or $y - 2 = \frac{2}{3}(m + 1 - \sqrt{m^2 - m + 1})(x - 1)$ and the line through $R(1, 2)$ and $E$ has an equation of $y - 2 = \frac{E_y - 2}{E_x - 1}(x - 1)$ or $y - 2 = \frac{2}{3}(m + 1 + \sqrt{m^2 - m + 1})(x - 1)$ .

Since $M$ is on $RD$ and $y = 2x + 2$ , $2M_x = \frac{2}{3}(m + 1 - \sqrt{m^2 - m + 1})(M_x - 1)$ , or $M_x = \frac{1 - \sqrt{m^2 - m + 1}}{m - 1}$ , and since $N$ is on $RE$ and $y = 2x + 2$ , $2N_x = \frac{2}{3}(m + 1 + \sqrt{m^2 - m + 1})(N_x - 1)$ , or $N_x = \frac{1 + \sqrt{m^2 - m + 1}}{m - 1}$ .

Since $M$ and $N$ are on the same line, the slope at which $|MN|$ reaches a minimum is equivalent to the slope at which $V = M_x - N_x$ is a minimum. Using substitution, $V = M_x - N_x = \frac{1 - \sqrt{m^2 - m + 1}}{m - 1} - \frac{1 + \sqrt{m^2 - m + 1}}{m - 1}$ or $V = -\frac{2\sqrt{m^2 - m + 1}}{m - 1}$ , which has a minimum when the derivative of $V$ is equal to $0$ (since $V'' > 0$ ), or when $V' = \frac{m + 1}{(m - 1)^2 \sqrt{m^2 - m + 1}} = 0$ , which solves to $m = \boxed{-1}$ .

Let $D=(t_1^2, 2t_1)$ and $E=(t_2^2, 2t_2)$ , $t_1, t_2 \ge 0$ . Then the equation of $DR$ is $\frac { y-2 }{ x-1 } =\frac { 2{ t }_{ 1 }-2 }{ { { t }_{ 1 } }^{ 2 }-1 } =\frac { 2 }{ { t }_{ 1 }+1 }$ $y=\frac { 2x+2{ t }_{ 1 } }{ { t }_{ 1 }+1 }$ $DR$ and $l$ intersect at $M$ : $2x+2=\frac { 2x+2{ t }_{ 1 } }{ { t }_{ 1 }+1 }$ $x=\frac{-1}{t_1}$ and $y=2(\frac { -1 }{ { t }_{ 1 } } )+2=2-\frac{2}{t_1}$ , we have $M=(\frac{-1}{t_1}, 2-\frac{2}{t_1})$ .

Similarly, $N=(\frac{-1}{t_2}, 2-\frac{2}{t_2})$ So $|MN|=\sqrt { (\frac { -1 }{ { t }_{ 1 } } -\frac { -1 }{ { t }_{ 2 } } )^ 2+((2-\frac { 2 }{ { t }_{ 1 } } )-(2-\frac { 2 }{ { t }_{ 2 } } ))^ 2 }$ Simplify, and WLOG, let $t_1<t_2$ , we get $|MN|=\sqrt{5}(\frac{1}{t_1}-\frac{1}{t_2})$ On the other hand, since $Q$ is on $DE$ , so $m_{DE}=\frac { 2{ t }_{ 1 }-2{ t }_{ 2 } }{ { { t }_{ 1 } }^{ 2 }-{ { t }_{ 2 } }^{ 2 } } =\frac { 2{ t }_{ 1 }-1 }{ { { t }_{ 1 } }^{ 2 }-1 }$ $\frac { 2 }{ { t }_{ 1 }+{ t }_{ 2 } } =\frac { 2{ t }_{ 1 }-1 }{ { { t }_{ 1 } }^{ 2 }-1 }$ $2t_1t_2-t_1-t_2+2=0$ Now we can use Lagrange Multiplier to do this problem: Let $L(t_1, t_2, \lambda)=\sqrt{5}(\frac{1}{t_1}-\frac{1}{t_2})-\lambda (2t_1t_2-t_1-t_2+2)$ Take the 3 partial derivatives, set them equal 0, we get $\begin{cases} (1)\quad \sqrt { 5 } (\frac { -1 }{ { { t }_{ 1 } }^{ 2 } } )-\lambda (2{ t }_{ 2 }-1)=0\quad \Rightarrow \quad \sqrt { 5 } +\lambda (2{ { t }_{ 1 } }^{ 2 }{ t }_{ 2 }-{ { t }_{ 1 } }^{ 2 })=0 \\ (2)\quad \sqrt { 5 } (\frac { 1 }{ { { t }_{ 2 } }^{ 2 } } )-\lambda (2{ t }_{ 1 }-1)=0\quad \Rightarrow \quad \sqrt { 5 } -\lambda (2{ t }_{ 1 }{ { t }_{ 2 } }^{ 2 }-{ { t }_{ 2 } }^{ 2 })=0 \\ (3)\quad -(2{ t }_{ 1 }{ t }_{ 2 }-{ t }_{ 1 }-{ t }_{ 2 }+2)=0\quad \Rightarrow \quad 2{ t }_{ 1 }{ t }_{ 2 }-{ t }_{ 1 }-{ t }_{ 2 }+2=0 \end{cases}$ (1)-(2), we get $\lambda (2{ { t }_{ 1 } }^{ 2 }{ t }_{ 2 }-{ { t }_{ 1 } }^{ 2 }+2{ t }_{ 1 }{ { t }_{ 2 } }^{ 2 }-{ { t }_{ 2 } }^{ 2 })=0$ $\lambda (2{ t }_{ 1 }{ t }_{ 2 }({ t }_{ 1 }+{ t }_{ 2 })-({ { t }_{ 1 } }^{ 2 }+{ { t }_{ 2 } }^{ 2 }))=0$ $\lambda (2{ t }_{ 1 }{ t }_{ 2 }({ t }_{ 1 }+{ t }_{ 2 })-{ ({ t }_{ 1 }+{ t }_{ 2 } })^{ 2 }+2{ t }_{ 1 }{ t }_{ 2 })=0$ By (3), we have $t_1+t_2=2t_1t_2+2$ , so the above equation simplify to $\lambda (2{ t }_{ 1 }{ t }_{ 2 }(2{ t }_{ 1 }{ t }_{ 2 }+2)-{ (2{ t }_{ 1 }{ t }_{ 2 }+2 })^{ 2 }+2{ t }_{ 1 }{ t }_{ 2 })=0$ $\lambda (-2t_1t_2-4)=0$ Obviously, $\lambda \neq 0$ (otherwise by (1), $\sqrt{5}=0$ ), so we must have $t_1t_2=-2$ and $t_1+t_2=2(-2)+2=-2$ Finally, recall that $m_{DE}=\frac{2}{t_1+t_2}=\frac{2}{-2}=\boxed{-1}$ Remark: One could use second derivative test to verify this indeed minimize $|MN|$