Solveeit Logo
Login

Question

Question: If the line \(ax+by+c=0,ab\ne 0\), is a tangent to the curve \(xy=1-2x\), then A. \(a>0,b<0\) B....

If the line ax+by+c=0,ab≠0ax+by+c=0,ab\ne 0, is a tangent to the curve xy=1−2xxy=1-2x, then
A. a>0,b<0a>0,b<0
B. a>0,b>0a>0,b>0
C. a<0,b>0a<0,b>0
D. a<0,b<0a<0,b<0

Explanation

Solution

Hint: We will be using the concept of differential calculus to find the slope of tangent and then we will be using point and slope form of line to write the equation of tangent after this we will compare the equation with the given one to find the answer.

Complete step-by-step solution -
Now, we have been given that ax+by+c=0,ab≠0ax+by+c=0,ab\ne 0 is a tangent to the curve xy=1−2xxy=1-2x.
Now, we know that tangent to a curve at any point is given by dydx or f′(x)\dfrac{dy}{dx}\ or\ f'\left( x \right). So, we differentiate xy=1−2xxy=1-2x with respect to x.
⇒xdydx+y=−2 ⇒xdydx=−2−y ⇒dydx=−2−yx..........(1) \begin{aligned} \Rightarrow & x\dfrac{dy}{dx}+y=-2 \\\ \Rightarrow & x\dfrac{dy}{dx}=-2-y \\\ \Rightarrow & \dfrac{dy}{dx}=\dfrac{-2-y}{x}..........\left( 1 \right) \\\ \end{aligned}
Now, let us suppose an arbitrary point (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) on the curve. Now we know that the line tangent through (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) is of the form,
(y−y1)=m(x−x1)\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)
Where m is the slope of the line. So, from (1) we have,
m=−2−y1x1 ⇒(y−y1)=−(2+y1)x1(x−x1) \begin{aligned} & m=\dfrac{-2-{{y}_{1}}}{{{x}_{1}}} \\\ \Rightarrow & \left( y-{{y}_{1}} \right)=\dfrac{-\left( 2+{{y}_{1}} \right)}{{{x}_{1}}}\left( x-{{x}_{1}} \right) \\\ \end{aligned}
Now, on cross multiply and simplifying, we have,
⇒x1y−x1y1=−2x+2x1−xy1+x1y1 ⇒x1y+x(2+y1)−2x1y1−2x1=0.........(2) \begin{aligned} \Rightarrow & {{x}_{1}}y-{{x}_{1}}{{y}_{1}}=-2x+2{{x}_{1}}-x{{y}_{1}}+{{x}_{1}}{{y}_{1}} \\\ \Rightarrow & {{x}_{1}}y+x\left( 2+{{y}_{1}} \right)-2{{x}_{1}}{{y}_{1}}-2{{x}_{1}}=0.........\left( 2 \right) \\\ \end{aligned}
Now, since (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) lies on the curve. Therefore, we have it satisfy the equation x1y1=1−2x1{{x}_{1}}{{y}_{1}}=1-2{{x}_{1}}.
x1(y1+2)=1{{x}_{1}}\left( {{y}_{1}}+2 \right)=1
Since, 1 > 0. So, we have,
x1(y1+2)>0...........(3){{x}_{1}}\left( {{y}_{1}}+2 \right)>0...........\left( 3 \right)
Now, we will compare (2) with ax+by+c=0ax+by+c=0. So, we have,
⇒ax+by+c=0 ⇒(2+y1)x+x1y−2x1y1−2x1=0 \begin{aligned} \Rightarrow & ax+by+c=0 \\\ \Rightarrow & \left( 2+{{y}_{1}} \right)x+{{x}_{1}}y-2{{x}_{1}}{{y}_{1}}-2{{x}_{1}}=0 \\\ \end{aligned}
So, on comparing we have,
⇒a=(2+y1).......(4) ⇒b=x1........(5) ⇒c=−2x1y1−2x1 \begin{aligned} \Rightarrow & a=\left( 2+{{y}_{1}} \right).......\left( 4 \right) \\\ \Rightarrow & b={{x}_{1}}........\left( 5 \right) \\\ \Rightarrow & c=-2{{x}_{1}}{{y}_{1}}-2{{x}_{1}} \\\ \end{aligned}
Now, from (3) we have,
x1(y1+2)>0{{x}_{1}}\left( {{y}_{1}}+2 \right)>0
This is only possible if (x1>0 and y1+2>0) or (x1<0 and y1+2<0)\left( {{x}_{1}}>0\ and\ {{y}_{1}}+2>0 \right)\ or\ \left( {{x}_{1}}<0\ and\ {{y}_{1}}+2<0 \right).
So, we have from (4) and (5)
a>0 and b>0 a<0 and b<0 \begin{aligned} & a>0\ and\ b>0 \\\ & a<0\ and\ b<0 \\\ \end{aligned}
So, the correct options are (B) and (D).

Note: To solve these types of questions it is important to remember the concepts of differential calculus and coordinate geometry. Also it is important to note that we have used the fact that if ab > 0 then either a > 0, b > 0 or a < 0, b < 0.