Question

If θ denotes the acute angle between the curves, y = 10 - x² and y = 2 + x² at a point of their intersection, then tanθ is equal to (A) 4/9
(B) 8/15
(C) 7/17
(D) 8/17
[JEE Main 2019]

Answer 1

If θ denotes the acute angle between the curves, then , tanθ = 8/15

• tanθ = $|\frac{m1 - m2}{1 + m1m2} |$,

Where,

• m1 = Slope of the first curve
• m2 = Slope of the second curve

First lets find the point of intersection of the two curves,

• y = 10 - $x^{2}$ and y = 2 +

• 10 - $x^{2}$ =  2 +
• 2$x^{2}$ = 8
• $x^{2}$ = 4
• x = ± 2
• y = 10 - $2^{2}$ = 6
• Hence the point of intersection of the 2 curves is (-2,6) and (+2,6)

Lets find the slope.

• m1 = $\frac{dy}{dx}$  = -2x ==> -4 at (2,6) and 4 at (-2,6)
• m2 = $\frac{dy}{dx}$ =  2x ==> 4 at (2,6) and -4 at (-2,6)

Consider (2,6)

• tanθ = $|\frac{m1 - m2}{1 + m1m2} |$, = $|\frac{-4 - 4 }{1 + (-4)(4)} |$  = $|\frac{-8}{-15} |$ = 8/15

(-2,6) will also give the same answer

Answer is option B