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Application of Derivatives
0
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Find the equation of the normal to curve \(y^2 = 4x\) which passes through the point \((1, 2).\)
cbse
class12
bookproblem
ch6
misc
q4
p242
sec-a
easy
math
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asked
Nov 28, 2012
by
thanvigandhi_1
retagged
Aug 17, 2013
by
sharmaaparna1
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1 Answer
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Toolbox:
$\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
The curve is $y^2=4x$
Differentiating with respect to $x$
$2y\large\frac{dy}{dx}$$=4$
$\large\frac{dy}{dx}=\frac{4}{2y}$
$\large\frac{dy}{dx}=\frac{2}{y}$
Step 2:
At $(1,2)$
$\large\frac{dy}{dx}=\frac{2}{2}$$=1$
Slope of normal $=-1$
Equation of normal at (1,2)
$y-2=(-1)(x-1)$
(or)$y-2=-x+1$
$x+y-3=0$
answered
Aug 9, 2013
by
sreemathi.v
edited
Sep 2, 2013
by
sharmaaparna1
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