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

Answer 1

Answer: (D)

$\frac{8}{15}$

Given curves $\Rightarrow y=10-x^2$

$y=2+x^2$

points of intersection

$10-x^2=2+x^2$

$\Rightarrow 2x^2=8$

$\Rightarrow x^2=4$

$\therefore x=\pm 2$

Co-ordinate $(\pm 2,6)$

Now the angle between two curves is angle between their tangents at that point.

for $y=2+x^2$

$\frac{dy}{dx}{|}_1=2x$

for $y=10-x^2$

$\frac{dy}{dx}{|}_2=-2x$

Now, as $\tan =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

(where $\theta$ is the angle between tangent)

$\Rightarrow \tan \theta =\left|\frac{4-(-4)}{1+4(-4)}\right|$

$\therefore \tan \theta =\frac{8}{15}$