Question

The equation of tangent to the curve y(1 + x²) = 2 - x, where it crosses x-axis is
(a) x + 5y = 2 (b) x - 5y = 2 (c) 5x - y = 2 (d) 5x + y = 2

Answer 1

The equation of given curve is

y(1 + x²) = 2 − x .....(1)

This cuts the x-axis at the point, where y = 0.

Putting y = 0 in (1), we get

0 × (1 + x²) = 2 − x

⇒ 2 − x = 0

⇒ x = 2

So, the given curve intersects the x-axis at (2, 0).

y(1 + x²) = 2 − x

Differentiating both sides with respect to x, we get

$y\times 2x+\left(1+x^2\right)\times \frac{dy}{dx}=-1$

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

$\Rightarrow \frac{dy}{dx}=-\frac{1+2xy}{1+x^2}$

∴ Slope of tangent to the given curve at (2, 0) = ${\left(\frac{dy}{dx}\right)}_{\left(2,0\right)}=-\frac{1+2\times 2\times 0}{1+{\left(2\right)}^2}=-\frac{1}{5}$

So, the equation of tangent at (2, 0) is

$y-0={\left(\frac{dy}{dx}\right)}_{\left(2,0\right)}\left(x-2\right)$

$\Rightarrow y=-\frac{1}{5}\left(x-2\right)$

$\Rightarrow 5y=-x+2$

$\Rightarrow x+5y=2$

Thus, the equation of tangent to the given curve where it crosses x-axis is x + 5y = 2.

Hence, the correct answer is option (a).