Question:
If the acute angle between the parabolas $y=8x-x^{2}$ and $y=x^{2}-4x$ is $\theta$, then $\tan\theta=$
If the acute angle between the parabolas $y=8x-x^{2}$ and $y=x^{2}-4x$ is $\theta$, then $\tan\theta=$
Show Hint
Always find the correct intersection coordinates first, then evaluate the numerical values of the derivative slopes at that point to plug into the tangent formula.
Updated On: Jun 3, 2026
- 1
- $\sqrt{3}$
- $\frac{15}{17}$
- $\frac{12}{31}$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Concept
The angle $\theta$ between two curves at their point of intersection is the angle between their tangents at that point, given by $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$, where $m_1$ and $m_2$ are the slopes of the curves at the intersection point.
Step 2: Meaning
First, find the intersection points by equating the curves: $8x - x^2 = x^2 - 4x \implies 2x^2 - 12x = 0 \implies 2x(x-6) = 0$. This gives $x = 0$ or $x = 6$. Let's analyze the non-trivial intersection point at $x = 6$.
Step 3: Analysis
Differentiate both curves to find their slope functions: 1. For $y_1 = 8x - x^2 \implies m_1 = \frac{dy_1}{dx} = 8 - 2x$. At $x = 6$, $m_1 = 8 - 12 = -4$. 2. For $y_2 = x^2 - 4x \implies m_2 = \frac{dy_2}{dx} = 2x - 4$. At $x = 6$, $m_2 = 12 - 4 = 8$.
Step 4: Conclusion
Substitute the slopes $m_1 = -4$ and $m_2 = 8$ into the angle formula: $\tan\theta = \left|\frac{-4 - 8}{1 + (-4)(8)}\right| = \left|\frac{-12}{1 - 32}\right| = \left|\frac{-12}{-31}\right| = \frac{12}{31}$. This perfectly matches option (D).
Final Answer: (D)
The angle $\theta$ between two curves at their point of intersection is the angle between their tangents at that point, given by $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$, where $m_1$ and $m_2$ are the slopes of the curves at the intersection point.
Step 2: Meaning
First, find the intersection points by equating the curves: $8x - x^2 = x^2 - 4x \implies 2x^2 - 12x = 0 \implies 2x(x-6) = 0$. This gives $x = 0$ or $x = 6$. Let's analyze the non-trivial intersection point at $x = 6$.
Step 3: Analysis
Differentiate both curves to find their slope functions: 1. For $y_1 = 8x - x^2 \implies m_1 = \frac{dy_1}{dx} = 8 - 2x$. At $x = 6$, $m_1 = 8 - 12 = -4$. 2. For $y_2 = x^2 - 4x \implies m_2 = \frac{dy_2}{dx} = 2x - 4$. At $x = 6$, $m_2 = 12 - 4 = 8$.
Step 4: Conclusion
Substitute the slopes $m_1 = -4$ and $m_2 = 8$ into the angle formula: $\tan\theta = \left|\frac{-4 - 8}{1 + (-4)(8)}\right| = \left|\frac{-12}{1 - 32}\right| = \left|\frac{-12}{-31}\right| = \frac{12}{31}$. This perfectly matches option (D).
Final Answer: (D)
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Angle Between Curves Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions