The condition that the line x/p+y/q = 1 be a normal to the parabola y² = 4ax is

Correct Answer is : p3p³p3 = 2ap2+aq22ap²+aq²2ap2+aq2

Correct Answer is : p3p³p3 = 2ap2+aq22ap²+aq²2ap2+aq2

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VITEEE 2017 Paper

Section: Mathematics

Question : 106 of 125

Marks: +1, -0

\frac{x}{p}+\frac{y}{q}

y^2

= 4ax is

Solution:

The line

\frac{x}{p}+\frac{y}{q}

= 1 will be a normal to the parabola

y^2

= 4ax if, for some value of m, it is identical with

y = mx - 2am -

am^3

i.e. mx - y = (2am +

am^3

)

Comparing coefficients, we get

\frac{m}{\frac{1}{p}}

\frac{-1}{\frac{1}{q}}

\frac{2a,m+am^3}{1}

⇒ mp = - q

∴ m =

-\frac{q}{p}

and mp = m (2a +

am^2

)

or P = 2a +

am^2

= 2a + a

\left(-\frac{q}{p}\right)^2

or p = 2a +

\frac{aq^2}{p^2}

p^3

2ap^2 + aq^2

Which is the required condition

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