Question

Find the equation of tangent of the curve \[y(1 + {x^2}) = 2 - x\] where it crosses the x-axis.

Answer 1

Hint: The equation of the curve is given, and it crosses the x-axis. First find the points and the first derivative. After equating both of them we will have the slope. And after putting the value we have the equation of tangent of the curve.

Complete step-by-step answer:
We have the equation of the given curve as \[y(1 + {x^2}) = 2 - x\].
This equation can be written after rearranging as $y = \dfrac{{2 - x}}{{1 + {x^2}}}$
To get the point where the curve will cross the x-axis, substitute the value of y as $y = 0$ in the given equation of the curve.
Therefore, we can get the value of x as follows:
 \[
  0 = \dfrac{{2 - x}}{{(1 + {x^2})}} \\
   \Rightarrow 0 = 2 - x \\
   \Rightarrow x = 2 \\
 \]
So, the given curve cuts the x- axis at the point (2, 0).
Now differentiating the given equation \[y(1 + {x^2}) = 2 - x\] with respect to x, using the quotient rule,
The quotient rule formula can be written as $\left( {d\left( {\dfrac{u}{v}} \right) = \dfrac{{v.d\left( u \right) - u.d\left( v \right)}}{{{v^2}}}} \right)$
We get the derivative of the given equation by solving as follows:
\[
  \Rightarrow y(0 + 2x) + (1 + {x^2}).\dfrac{{dy}}{{dx}} = 0 - 1 \\
 \Rightarrow 2xy + (1 + {x^2})\dfrac{{dy}}{{dx}} = - 1 \\
    \\
 \]
On simplifying it further, we get:
\[
  \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1 - 2xy}}{{1 + {x^2}}} \\
    \\
 \]
We know that \[\dfrac{{dy}}{{dx}}\] is the slope of the tangent.
So the slope of a tangent to this curve at (2,0) is
\[
 \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{(2,0)}} = \dfrac{{ - 1 - 2.2.0}}{{1 + {2^2}}} \\
\Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{(2,0)}} = \dfrac{{ - 1}}{5} \\
 \]
Now, the Equation of tangent of the curve passing through (2, 0) can be calculated as follows:
$
\Rightarrow y - 0 = - \dfrac{1}{5}(x - 2) \\
\Rightarrow 5y + x = 2 \\
 $

$5y+x=2$ is the equation of the tangent of the curve.

Note:
X- axis is a point on the graph where the y intercept is zero.
Similarly, y- axis is a point on the graph where the x intercept is zero.
As the curve crosses the x- axis, so we have the value of y as $y = 0$.
The equation of the tangent is calculated by the formula $y - {y_1} = \dfrac{{dy}}{{dx}}(x - {x_1})$