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KEAM 2013 Math Paper

Question : 120 of 120

Marks: +1, -0

r

with centres on the y-axis, are respectively

Solution:

The equation of family of circles of fixed radius'

r'

with centres on the

y

-axis is

(x-0)^{2}+(y-a)^{2}=r^{2}

.....(i)

On differentiating w.r.t.

x

, we get

2x+2(y-a)\;\frac{dy}{dx}=0

\Rightarrow \;\;\;\frac{dy}{dx}=-\frac{x}{y-a}

\Rightarrow \;\; (y-a)=\frac{-x}{\frac{dy}{dx}}

On putting this value in Eq. (i), we get

x^{2}+\frac{x^{2}}{\left(\frac{dy}{dx}\right)^{2}}=r^{2}

\Rightarrow \;\; x^{2}\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}=r^{2}\left(\frac{dy}{dx}\right)^{2}

Hence, order

\rightarrow

highest order derivative

=1

and degree

\rightarrow

power of highest order derivative

=2

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