Question

Angle between two curves $y^2=4(x+1)$ and $x^2=4(y+1)$ is

• A. $0^0$
• B. ${90}^0$
• C. ${60}^0$
• D. ${30}^0$

Answer 1

The correct option is A $0^0$

We have to find the area between the two curves $y^2=4(x+1),x^2=4(y+1)$

The graph of the two curves and their intersection points are shown below

These two curves intersects at two points $(4.828,4.828)$ and $(-0.828,-0.828)$.

Now find $\frac{dy}{dx}$ for both the equations at $(4.828,4.828)$

For $x^2=4(y+1),\frac{dy}{dx}=\frac{x}{2}$

At $(4.828,4.828)$, $\frac{dy}{dx}=\frac{4.828}{2}=2.414=s_1\left(say\right)$

For $y^2=4(x+1),\frac{dy}{dx}=\frac{x}{2}$(by implicit differentiation)

At $(4.828,4.828)$, $\frac{dy}{dx}=\frac{4.828}{2}=2.414=s_2\left(say\right)$

Now $tan\theta =\left|\frac{s_1-s_2}{1-s_1{s}_2}\right|$

Since $s_1=s_2$, they cut at $0^{\circ }$ angle, actually they touch each other.

Thus $\theta =0^{\circ }$