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VITEEE 2010 Paper

Question : 101 of 120

Marks: +1, -0

lx+my=1

y^{2}=4ax,

Solution:

Let

P(x_{1},y_{1})

be a point on the curve

y^{2}=4ax

...(i)

On differentiating

y^{2}=4ax

w.r.t. 'x', we get

2y\frac{dy}{dx}=4a

⇒

\left(\frac{dy}{dx}\right)_{(x_{1},y_{1})}=\frac{2a}{y_{1}}

Thus, the equation of normal at

(x_{1},y_{1})

y-y_{1}=\frac{y_{1}}{2a}(x-x_{1})

⇒

y_{1}x+2ay=y_{1}(x_{1}+2a)

...(ii)

But

lx+my=1

...(iii)

is also a normal.

Therefore, coefficients of eqs. (ii) and (iii), must be proportional.

i.e.,

\frac{y_{1}}{l}=\frac{2a}{m}=\frac{y_{1}(x_{1}+2a)}{1}

\Rightarrow y_{1}=\frac{2a l}{m}

and

x_{1}=\frac{1}{l}-2a

Putting these values of

x_{1}\text{ and }y_{1}

in eq. (i), we get

\left(\frac{2a l}{m}\right)^{2}=4a\left(\frac{1}{l}-2a\right)

⇒

\frac{4a^{2}l^{2}}{m^{2}}=\frac{4a-8a^{2}l}{l}

⇒

a l^{3}=m^{2}-2a m^{2}l\Rightarrow a l^{3}+2a l m^{2}=m^{2}

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