Question

If xy = 4 and x^2 =4y Then the common tangent is?
​

Answer 1

Given :

The equations $xy=4$ and $x^{2}=4y$

To find :

The common targent

Solution :

Let the tangent be $y=mx+c$

Since this tangent touches the curve $x^{2}=4y$, we can write

$\quad x^{2}=4(mx+c)$

$\Rightarrow x^{2}-4mx-4c=0$ .....(1)

Since the line $y=mx+c$ touches the parabola $x^{2}=4y$, then the two points of intersection will be coincident, that is, the roots of (1) are equal.

The condition for equal roots is

$\quad (-4m)^{2}-4(1)(-4c)=0$

$\Rightarrow 16m^{2}+16c=0$

$\Rightarrow m^{2}+c=0$

$\Rightarrow c=-m^{2}$

Putting $c=-m^{2}$ in the equation of the tangent, we get

$\quad y=mx-m^{2}$ .....(2)

Again since (2) is tangent to the rectangular hyperbola $xy=4$, we can write

$\quad x(mx-m^{2})=4$

$\Rightarrow mx^{2}-m^{2}x-4=0$ .....(3)

Since the line $y=mx-m^{2}$ touches the rectangular hyperbola $xy=4$, then the two points of intersection will be coincident, that is, the roots of (3) are equal.

The condition for equal roots is

$\quad (-m^{2})^{2}-4(m)(-4)=0$

$\Rightarrow m^{4}+16m=0$

$\Rightarrow m(m^{3}+16)=0$

This gives either $m=0$ or $m=-2\sqrt[3]{2}$

When $m=0$, $c=0$ and the tangent line obtained is $y=0$, a common tangent for sure, which is verified by the added graph.

When $m=-2\sqrt[3]{2}$, $c=4\sqrt[3]{4}$ and the tangent line obtained is $y=-2\sqrt[3]{2}x+4\sqrt[3]{4}$, not a common tangent, which is verified by the added graph.

• [ Refer to the added picture. ]

Answer :

• $y=0$ i.e. the x -axis is the common tangent to the given curves $xy=4$ and $x^{2}=4y$.

Answer 2

Answer:

Tangent to x2+y2=4

eq. of tangent to circle is y=mx±21+m2.....1

x2=4y

Eq. 1 is also tangent to the parabola 

x2=4mx±81+m2

x2−4mx∓81+m2=0

D=0

16m2±4.81+m2=0

m2±21+m2=0

m2=±21+m2

m4=4+4m2

m4−4m2−4=0

m2=24±16+16

m2=2+2+2