SATHEE: Application of Derivatives 1 Question 6

Application of Derivatives 1 Question 6

####6. If $θ$ denotes the acute angle between the curves, $y = 10 - x^{2}$ and $y = 2 + x^{2}$ at a point of their intersection, then $| \tan θ |$ is equal to

(2019 Main, 9 Jan I)

(a) $\frac{7}{17}$

(b) $\frac{8}{15}$

(d) $\frac{8}{17}$

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Answer:

Correct Answer: 6. (b)

Solution:

Key Idea Angle between two curves is the angle between the tangents to the curves at the point of intersection.

Given equation of curves are

and $\begin{aligned} y & = 10 - x^{2} \\ y & = 2 + x^{2} \end{aligned}$

For point of intersection, consider

$\begin{aligned} 10 - x^{2} & = 2 + x^{2} \\ \Rightarrow & 2 x^{2} & = 8 \\ \Rightarrow & x^{2} & = 4 \\ \Rightarrow & x & = \pm 2 \end{aligned}$

Clearly, when $x = 2$ , then $y = 6$ (using Eq. (i)) and when $x = - 2$ , then $y = 6$

Thus, the point of intersection are $(2, 6)$ and $(- 2, 6)$ .

Let $m_{1}$ be the slope of tangent to the curve (i) and $m_{2}$ be the slope of tangent to the curve (ii)

For curve (i) $\frac{d y}{d x} = - 2 x$ and for curve (ii) $\frac{d y}{d x} = 2 x$

$∴$ At $(2, 6)$ , slopes $m_{1} = - 4$ and $m_{2} = 4$ , and in that case

$| \tan θ | = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} | = | \frac{4 + 4}{1 - 16} | = \frac{8}{15}$

At $(- 2, 6)$ , slopes $m_{1} = 4$ and $m_{2} = - 4$ and in that case

$| \tan θ | = | \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} | = | \frac{- 4 - 4}{1 - 16} | = \frac{8}{15}$

Application of Derivatives 1 Question 6

Answer:

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Application of Derivatives 1 Question 6

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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