Question

The equation of the common tangent to the curves, ${{y}^{2}}=16x$ and $xy=-4$ is A. $ x+y+4=0 $
B. $x-2y+16=0$ C. $ 2x-y+2=0 $
D. $x-y+4=0$ $$$$

Answer 1

The slope of the tangent at any point is the slope of the curve at point. See that the first equation is an equation of parabola of the form ${{y}^{2}}=4ax$ whose tangent equation we know as $y=mx+\dfrac{a}{m}$ where $m$ is the slope. Find the value of $a$ from the first equation and put $y=mx+\dfrac{a}{m}$ in the second equation. Simplify to obtain a quadratic equation whose roots are possible values of $m$ .

Complete step-by-step answer:
The given equation of curves is

& {{y}^{2}}=16x...\left( 1 \right) \\\ & xy=-4.....\left( 2 \right) \\\ \end{aligned}$$ The general equation of any parabola is $ {{y}^{2}}=4ax $ and The general equation of a rectangular hyperbola is $ xy=c $ where $ c $ is a real constant. The equation(1) is an equation of parabola and equation(2) is an equation of hyperbola.$$$$ We know from differential callus that calculus that the slope of any curve is given by the differentiation with respect to the independent variable. We also know that the slope of the curve at any point is the slope of the tangent at that point. As curves (1) and (2) have a common tangent we assign its slope as $ m. $ $$$$ The general equation of any parabola is $ {{y}^{2}}=4ax $ . Comparing it with equation (1) we get$${{y}^{2}}=16x=4\left( 4 \right)x$$ . The general equation of a rectangular hyperbola is $ xy=c $ where $ c $ is a real constant . Comparing it with equation(2) we get $ c=-4 $ $$$$ The equation of tangent at any point on the parabola is given by $$y=mx+\dfrac{a}{m}$$ We deduce from the equation(1) that $${{y}^{2}}=4\cdot 4x\Rightarrow a=4$$. So the equation of the tangent transforms to $$y=mx+\dfrac{4}{m}...(3)$$ Putting this equation in the equation (2) , we get $$\begin{aligned} & x\left( 4x+\dfrac{4}{m} \right)=-4 \\\ & \Rightarrow {{m}^{2}}{{x}^{2}}+4x+4m=0 \\\ \end{aligned}$$ The above equation is a quadratic equation in one variable whose root is the slope of the common tangent . So the root is unique and real. Then the discriminant of the standard quadratic equation for unique and real roots is $ a{{x}^{2}}+bx+c=0 $ is $ D={{b}^{2}}-4ac=0 $ . Using this fact, $$\begin{aligned} & \therefore D={{4}^{2}}-4{{m}^{2}}\cdot m=0 \\\ & \Rightarrow {{m}^{3}}=1 \\\ & \Rightarrow m=1 \\\ \end{aligned}$$ We have rejected other values of $ m $ as $ m\in R $ . Putting in equation(3) we get $$\begin{aligned} & y=x+4 \\\ & \Rightarrow y-x+4=0 \\\ \end{aligned}$$ **So, the correct answer is “Option D”.** **Note:** We chose to start with a tangent equation of parabola that is $ y=mx+\dfrac{a}{m} $ because it is easier to put in the equation of rectangular hyperbola. If we try to first find the tangent equation of rectangular hyperbola $ XY=C\Rightarrow \left( x+y \right)\left( x-y \right)=C\Rightarrow \dfrac{{{x}^{2}}}{C}-\dfrac{{{y}^{2}}}{C}=1 $ where $ C={{a}^{2}}={{b}^{2}} $ , we will get two tangents $ y=mx\pm \sqrt{C\left( {{m}^{2}}-1 \right)} $ . It will be cumbersome when we put them in the equation of parabola.