Question

The curve $xy=a^2$ and $x^2+y^2=2a^2$ touch each other.

• A. True
• B. False

Answer 1

The correct option is A True

The given curves are

$xy=a^2$........(1)

$x^2+y^2=2a^2$..........(2)

Now,

$(x+y{)}^2=x^2+y^2+2xy=2a^2+2a^2$

$\therefore (x+y{)}^2=4a^2$

$\Rightarrow x+y=2a,-2a$..........(3)

Also, $(x-y{)}^2=x^2+y^2-2xy$

$=2a^2-2a^2=0$

$\therefore x-y=0$...........(4)

Adding 3 and 4, we get,

$2x=2a,-2a$

$x=a,-a$

$\therefore Fromy,y=a,-a$

$\therefore$ Curves 1 and 2 intersect at $(a,a)$ and $(-a,-a)$

From $1,y=\frac{a^2}{x}$

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

From $2,2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$

Let $m_1,m_2$ be slopes of curve $1$ and $2$

At $(a,a),m_1=\frac{-a^2}{a^2}=-1$ and $m_2=\frac{-a}{a}-1$

$\therefore$ $At(a,a)$ curve touch.

At $(-a,-a),m_1=\frac{-a^2}{a^2}=-1$ and $m_2=\frac{-a}{-a}-1$

$\therefore$ At $(-a,-a)$ curve touch.