Prove that the curves $x = y^{2}$ and $x y = k$ cut at right angles if $8 k^{2} = 1$ .

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Solution

$x = y^{2}; x y = k$
Let the curves cut each other at $(a, b)$
$a = b^{2}; a b = k$
$x = y^{2}; x y = k$
$1 = 2 y \frac{d y}{d x}; x \frac{d y}{d x} + y = 0$
$\frac{d y}{d x} = \frac{1}{2} y; \frac{d y}{d x} = \frac{- y}{x}$
Slope at $a, b = \frac{1}{2 b}$ ; Slope at $a, b = \frac{- b}{a}$
$m_{1} = \frac{1}{2 b}; m_{2} = \frac{- b}{a}$
If the curves cut at right angles
$m_{1} m_{2} = - 1$
$\frac{1}{2 b} \frac{- b}{a} = - 1$
$2 a = 1 \to 1$
$a = b^{2}; a b = k$
$b^{2} . b = k$
$b^{3 / 2} = k$
$b = k^{2 / 3}$
Substituting in eq. $1$
$2. b^{2} = 1$
$2. k^{2 / 3} = 1$
$∴ 8 k^{2} = 1$

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Q. Prove that the curves

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x y = k

cut at right angles, if

8 k^{2} = 1

Prove that the curves $x = y^{2}$ and xy=k cut at right angle if $8 k^{2} = 1$ .

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x = y^{2} and xy = k

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Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1. [Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.]

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Prove that the curves x=y2 and xy=k cut at right angles if 8k2=1.

x=y2;xy=kLet the curves cut each other at (a,b)a=b2;ab=kx=y2;xy=k1=2ydydx;xdydx+y=0dydx=12y;dydx=−yxSlope at a,b=12b; Slope at a,b=−bam1=12b;m2=−baIf the curves cut at right anglesm1m2=−112b−ba=−12a=1→1a=b2;ab=kb2.b=kb3/2=kb=k2/3Substituting in eq. 12.b2=12.k2/3=1∴8k2=1

Prove that the curves $x = y^{2}$ and $x y = k$ cut at right angles if $8 k^{2} = 1$ .