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The line $y=x+1$ is a tangent to the curve $y^2=4 x$ at the point
(a) $(1,2)$
(b) $(2,1)$
(c) $(1,-2)$
(d) $(-1,2)$
The line $y=x+1$ is a tangent to the curve $y^2=4 x$ at the point
(a) $(1,2)$
(b) $(2,1)$
(c) $(1,-2)$
(d) $(-1,2)$
Solution:
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Verified Answer
(a) The curve is $y^2=4 x, \therefore \frac{d y}{d x}=\frac{4}{2 y}=\frac{2}{y}$
Slope of the given line $y=x+1$ is $1 . \therefore \frac{2}{y}=1$
$\therefore \quad y=2 \quad$ Putting $y=2$ in $y^2=4 x \quad 2^2=4 x$
$\Rightarrow \quad \mathrm{x}=1 \quad \therefore$ Point of contact is $(1,2)$
Slope of the given line $y=x+1$ is $1 . \therefore \frac{2}{y}=1$
$\therefore \quad y=2 \quad$ Putting $y=2$ in $y^2=4 x \quad 2^2=4 x$
$\Rightarrow \quad \mathrm{x}=1 \quad \therefore$ Point of contact is $(1,2)$
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