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

y = 2 x

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x + 2 y = 5

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2 x + y = 4

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None of these

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Solution

The correct option is C $y = 2 x$
$y = x^{2} + 1$
$\frac{d y}{d x} = 2 x$
Thus slope of tangent at $(1, 2)$ is, $m = 2 \times 1 = 2$
Therefore, the required tangent is given by,
$(y - 2) = 2 (x - 1) \Rightarrow y = 2 x$
Hence, option 'A' is correct.

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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} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0

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