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AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 69 of 160

Marks: +1, -0

If the line

ax+by+c=0

is a normal to the curve

x y=1

, then

$a >0, b>0$
$a>0, b< 0$
$a>0, b=0$
$a< 0, b< 0$

Solution: 👈: Video Solution

Let the curve and the normal intersect at point

(x_1, y_1)

x_1 y_1=1

and

a x_1+b y_1+c=0

Differentiating

x y=1

w.r.t. '

x

',

x \;\frac{dy}{dx}+y=0

\Rightarrow \left. \frac{dy}{dx} \right|_{(x_1, y_1)} =\; \frac{-y_1}{x_1}

\Rightarrow -\; \left. \frac{dx}{dy} \right|_{(x_1, y_1)} =\; \frac{x_1}{y_1}

Equation of normal is

y-y_1=\; \frac{x_1}{y_1} (x-x_1)

\Rightarrow y y_1-y_1^2=x x_1-x_1^2 \Rightarrow x x_1-y y_1=x_1^2-y_1^2

The product of

x_1 y_1

is positive,

x_1 y_1

are of same sign.

\Rightarrow a

and

b

are of opposite sign.

From options, we can conclude that

a>0, b< 0

.

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