Question

Show that the curves $xy=a^2$ and $x^2+y^2=2a^2$ touch each other.

Answer 1

Given equation of the curves $xy=a^2\cdots \left(1\right)$ and $x^2+y^2=2a^2\cdots \left(2\right)$

From $\left(1\right)$

$y=\frac{a^2}{x}\cdots \left(3\right)$

From $\left(2\right)$

$⟹x^2+y^2=2a^2$

$⟹x^2+{\left(\frac{a^2}{x}\right)}^2=2a^2$

$⟹x^2+\frac{a^4}{x^2}=2a^2$

$⟹x^4-2a^2{x}^2+a^4=0$

$⟹(x^2-a^2{)}^2=0$ $(\because (a-b{)}^2=a^2-2ab+b^2)$

$⟹x^2-a^2=0$

$⟹x=\pm a$

$y=\frac{a^2}{x}=\pm a$

So point of intersection of both the curves is $(a,a)$ and $(-a,-a)$

From $\left(3\right)$

$\frac{dy}{dx}=-\frac{a^2}{x^2}$

From $\left(2\right)$

$2x+2y\frac{dy}{dx}=0⟹\frac{dy}{dx}=-\frac{x}{y}$

Slope of the tangent to curve $xy=a^2$ at $(a,a)$ is $m_1=\frac{dy}{dx}{|}_{(a,a)}=-\frac{a^2}{a^2}=-1$

Slope of the tangent to curve $x^2+y^2=2a^2$ at $(a,a)$ is $m_2=\frac{dy}{dx}{|}_{(a,a)}=-\frac{a}{a}=-1$

As we know that

If the angle between two lines with slopes $m_1,m_2$ is $\theta$ then $\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$

Let the angle between the tangents be $\theta$

$⟹\tan \theta =\frac{-1+1}{1+(-1)(-1)}=0$

$⟹\theta =0$

Angle between the tangents is zero

So both tangents represents same line

Hence the curves $xy=a^2$ and $x^2+y^2=2a^2$ touch each other