Question

The equation of tangent to the curve y(1+x²)=2-x, where it crosses x axis is

Answer 1

Answer:

$\bold\red{x+5y-2=0}$

Step-by-step explanation:

Given,

A curve having equation,

$y(1 + {x}^{2} ) = 2 - x \\ \\ = > y + y {x}^{2} = 2 - x$

Now,

to find the Equation of tangent,

we need to differentiate the Equation,

Therefore,

we get,

$= > \frac{dy}{dx} + 2xy + {x}^{2} \frac{dy}{dx} = - 1 \\ \\ = > (1 + {x}^{2} ) \frac{dy}{dx} = - 1 - 2xy \\ \\ = > \frac{dy}{dx} = - \frac{(1 + 2xy)}{(1 + {x}^{2} )}$

Now,

it is crossing the X axis,

Thus,

Put y = 0

Therefore,

we get,

$= > 2 - x = 0 \\ \\ = > x = 2$

Therefore,

The curve intersected the X axis at (2,0)

Therefore,

slope at this point is,

$= > \frac{dy}{dx} = - \frac{1}{1 + {2}^{2} } = - \frac{1}{5}$

Therefore,

equation of tangent is,

$= > \frac{y - 0}{x - 2} = - \frac{1}{5} \\ \\ = > 5y = 2 - x \\ \\ = > x + 5y - 2 = 0$

Hence,

Equation of tangent is $\bold{x+5y-2=0}$