If y2-4ax- then 1-y123-2-y2 at x-a is

The correct option is A 4√(2)a 2y #160; y^1=4a 2(y^1)^2+2y y^11=0 (1+y_1^2)^3/2y_2= ( 1+ ( 4a2y )^2 )^3/2 2y^122y [ y=2a amp ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If y2=4ax, ...

Question

If $y^{2} = 4 a x$ , then $\frac{(1 + y_{1}^{2})^{3 / 2}}{y_{2}}$ at $x = a$ is

- 4 \sqrt{2} a

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4 \sqrt{2} a

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\frac{4 \sqrt{2}}{a}

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- 4 a

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Solution

The correct option is A $- 4 \sqrt{2} a$
$2 y y^{1} = 4 a$
$2 (y^{1})^{2} + 2 y y^{11} = 0$
$\frac{(1 + y_{1}^{2})^{3 / 2}}{y_{2}} = \frac{{(1 + {(\frac{4 a}{2 y})}^{2})}^{3 / 2}}{\frac{- 2 y^{12}}{2 y}}$
$(\begin{matrix} y = 2 a & a t x = a y^{1} = 1 & a t x = a \end{matrix}) = - (1 + 1)^{3 / 2} (2 a)$
$= - 4 \sqrt{2} a$

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If y2=4ax, then (1+y21)3/2y2 at x=a is

The correct option is A −4√2a2yy1=4a2(y1)2+2yy11=0(1+y21)3/2y2=(1+(4a2y)2)3/2−2y122y(y=2aatx=ay1=1atx=a)=−(1+1)3/2(2a)=−4√2a

If $y^{2} = 4 a x$ , then $\frac{(1 + y_{1}^{2})^{3 / 2}}{y_{2}}$ at $x = a$ is