Find the equation of the tangent to the curve $y = x^{2} + 1$ at the point $(1, 2) .$

2 y = x

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y = 2 x

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y + x = 2

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y + 2 x = 0

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Solution

The correct option is C $y = 2 x$
Given $y = x^{2} + 1$

$y^{'} = 2 x$ at point $(1, 2)$
$\Rightarrow y^{'} = 2$

The equation of tangent $y - 2 = 2 (x - 1)$
$\Rightarrow y = 2 x$
Ans: B

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List-I	List-II
a) Equation of tangent to the curve $y = b e^{- x / a}$ at $x = 0$	1) $x - 2 y = 2$
b) Equation of tangent to the curve $y = x^{2} + 1$ at $(1, 2)$	2) $y = 2 x$
c) Equation of normal to the curve $y = 2 x - x^{2}$ at $(2, 0)$	3) $x - y = π$
d) Equation of normal to the curve $y = sin x$ at $x = π$	4) $\frac{x}{a} + \frac{y}{b} = 1$

x^{2} + y^{2} - 2 x + 4 y = 0

y = x^{2} + 1

(1, 2)

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