Question

If $\theta$ denotes the acute angle between the curves, $y=10-x^2$ and $y=2+x^2$ at a point of their intersection, then $|\tan \theta |$ is equal to :

• A. $\frac{8}{17}$
• B. $\frac{8}{15}$
• C. $\frac{4}{9}$
• D. $\frac{7}{17}$

Answer 1

The correct option is B $\frac{8}{15}$

Given :
The curves
$y=10-x^2$.........(1)
$y=2+x^2$............(2)

Finding the point of intersection of the two curves,
$10-x^2=2+x^2$
$\Rightarrow 2x^2=8\Rightarrow x=\pm 2$
So, the point of intersection will be $(2,6)\text{and}(-2,6)$.

The angle between the curves is the angle between their tangents at the point of intersection.
For point of intersection $(2,6)$
Slope of tangent of curve (1) ; $m_1=-2x=-4$
Slope of tangent of curve (2) ; $m_2=2x=4$
$\therefore |\tan \theta |=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{-8}{1-16}\right|=\frac{8}{15}$

For point of intersection $(-2,6)$
Slope of tangent of curve (1) ; $m_1=-2x=4$
Slope of tangent of curve (2) ; $m_2=2x=-4$
$\therefore |\tan \theta |=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{8}{1-16}\right|=\frac{8}{15}$

Hence, $|\tan \theta |=\frac{8}{15}$