KCET 2016 Math Solved Paper

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