Question

The tangent of the acute angle between the curves $y=|x^2-1|$ and $y=\sqrt{7-x^2}$ at their points of intersection is

• A. $\frac{5\sqrt{3}}{2}$
• B. $\frac{3\sqrt{5}}{2}$
• C. $\frac{5\sqrt{3}}{4}$
• D. $\frac{3\sqrt{5}}{4}$

Answer 1

The correct option is C $\frac{5\sqrt{3}}{4}$

The given curves are $y=|x^2-1|$ and $y=\sqrt{7-x^2}$

To find the point of intersection, let us equate both the curves.

$|x^2-1|=\sqrt{7-x^2}$

Squaring on both sides gives us $x^4-x^2-6=0$

$(x^2-3)(x^2+2)=0$

On solving above quadratic equation, we get $x^2=3$

The point of intersection is $(\pm \sqrt{3},2)$

The slope of the tangent to $y=|x^2-1|$ at the point $(\sqrt{3},2)$is

$m_1=\frac{dy}{dx}=2x=2\sqrt{3}$

The slope of tangent to $y=\sqrt{7-x^2}$ at the point $(\sqrt{3},2)$is

$m_2=\frac{dy}{dx}=\frac{-x}{\sqrt{7-x^2}}=\frac{-\sqrt{3}}{2}$

Let $\theta$ be the acute angle between the two tangents.

$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$

$\Rightarrow \tan \theta =\frac{-5\sqrt{3}}{4}$

$\therefore$The tangent of the acute angle between the curves is $\frac{5\sqrt{3}}{4}$ .

Hence, option C is correct.