The slope of normal at any point P of a curve (lying in the first quadrant) is reciprocal of twice the product of the abscissa and the ordinate of point P. Then, the equation of the curve is (where, c is an arbitrary constant)

Question

The slope of normal at any point P of a curve (lying in the first quadrant) is reciprocal of twice the product of the abscissa and the ordinate of point P. Then, the equation of the curve is (where, c is an arbitrary constant) (A) y2=x+c (B) y = c e- x2 (C) y=ce- x (D) y2=ln x+c

Tardigrade Admin · Accepted Answer

Slope of normal at P=(- 1/((d y)d x)P) ⇒ (1/2 x y)=(- 1/((d y)d x)) or (d y/d x)=-2xy ⇒ displaystyle ∫ (d y/y)= displaystyle ∫ - 2 xdx ⇒ ln y=-x2+k ⇒ y = e- x2 + k = c e- x2

Q. The slope of normal at any point $P$ of a curve (lying in the first quadrant) is reciprocal of twice the product of the abscissa and the ordinate of point $P .$ Then, the equation of the curve is (where, $c$ is an arbitrary constant)

$y^{2} = x + c$

$y = c e^{- x^{2}}$

$y = c e^{- x}$

$y^{2} = l n x + c$

Slope of normal at $P = \frac{- 1}{( \frac{d y}{d x} ) _{P}}$
$\Rightarrow \frac{1}{2 x y} = \frac{- 1}{( \frac{d y}{d x} )}$
or $\frac{d y}{d x} = - 2 x y$
$\Rightarrow \int \frac{d y}{y} = \int - 2 x d x$
$\Rightarrow l n y = - x^{2} + k$
$\Rightarrow y = e^{- x^{2} + k} = c e^{- x^{2}}$

Q. The slope of normal at any point P of a curve (lying in the first quadrant) is reciprocal of twice the product of the abscissa and the ordinate of point P. Then, the equation of the curve is (where, c is an arbitrary constant)

y2=x+c

y=ce−x2

y=ce−x

y2=lnx+c