Given curve, y(1+x2)=2−x → (1) At x-axis, y = 0 so Eq. (1) becomes ⇒0=2−x⇒x=2 So, the point is (2, 0). Now, differentiate Eq. (1) with respect to x, we get y′(1+x2)+y(2x)=−1 At point (2, 0), we have ⇒y′(1+22)+0(2×2)=−1 ⇒y′(5)=−1⇒y′=
−1
5
This is the slope of the tangent at point (2, 0). So, slope of normal is −
1
y′
=5 Therefore, equation of normal is, y−0=5(x−2) ⇒y=5x−10 ⇒5x−y−10=0