The maximum area of a triangle whose one vertex is at \( (0, 0) \) and the other two vertices lie on the curve \( y = -2x^2 + 54 \) at points \( (x, y) \) and \( (-x, y) \) where \( y > 0 \) is:
- 88
- 108
- 92
- 122
The Correct Option is B
Approach Solution - 1
To find the maximum area of the triangle with vertices at (0, 0), (x, y), and (−x, y) where \( y = -2x^2 + 54 \), we can proceed as follows:
The base of the triangle is the distance between (x, y) and (−x, y), which is 2x.
The height of the triangle is y, which is the distance from the origin (0, 0) to the line joining (x, y) and (−x, y).
Thus, the area \( \Delta \) of the triangle is:
\[\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2x \times y = x \times y\]
Since \( y = -2x^2 + 54 \), we can substitute this into the area expression:
\[\Delta = x \times (-2x^2 + 54) = -2x^3 + 54x\]
To maximize \( \Delta \), we take the derivative with respect to \( x \) and set it to zero:
\[\frac{d\Delta}{dx} = -6x^2 + 54 = 0\]
\[-6x^2 = -54 \Rightarrow x^2 = 9 \Rightarrow x = 3 \quad (\text{since } y > 0) \]
Now, substitute \( x = 3 \) back into the equation for \( y \):
\[ y = -2(3)^2 + 54 = -18 + 54 = 36 \]
Thus, the maximum area is:
\[ \Delta = x \times y = 3 \times 36 = 108 \]
Approach Solution -2
The problem involves finding the maximum area of a triangle with a vertex at the origin \((0, 0)\) and the other two vertices on the parabola defined by \(y = -2x^2 + 54\). The symmetry in the problem arises because the vertices are \((x, y)\) and \((-x, y)\), meaning the triangle's base is along the x-axis, and its height is perpendicular to it.
- Given the curve \(y = -2x^2 + 54\), substitute the coordinates of the two vertices into this equation to maintain their alignment on the curve.
- The area \(A\) of the triangle with a vertex at the origin and a base along the x-axis will be given by: \(A = \frac{1}{2} \times \text{base} \times \text{height}\).
- Here, the base is the horizontal distance between \((x, 0)\) and \((-x, 0)\), which is \(2x\), and the height is \(y\). \(A = \frac{1}{2} \times 2x \times y = x \times y\).
- Substitute for \(y\): \(y = -2x^2 + 54\), giving: \(A = x(-2x^2 + 54) = -2x^3 + 54x\).
- To find the maximum area, differentiate \(A\) with respect to \(x\) and set it to zero: \(\frac{dA}{dx} = -6x^2 + 54 = 0\). Solving gives: \(-6x^2 = -54 \Rightarrow x^2 = 9 \Rightarrow x = 3 \text{ or } x = -3\).
- Choose \(x = 3\) (since symmetry and maximum areas are positive). Compute \(y\) at this point using: \(y = -2(3)^2 + 54 = -18 + 54 = 36\).
- The area \(A\) at \(x = 3\) is: \(A = 3 \cdot 36 = 108\).
Thus, the maximum area of the triangle is 108 square units.
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Area of a Triangle - by Heron’s Formula Questions
- The area of the region bounded by the curves \( x(1 + y^2) = 1 \) and \( y^2 = 2x \) is:
Let \( A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix} \) and \( P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \theta > 0. \) If \( B = P A P^T \), \( C = P^T B P \), and the sum of the diagonal elements of \( C \) is \( \frac{m}{n} \), where gcd(m, n) = 1, then \( m + n \) is:
- Let \( f \) be a real-valued continuous function defined on the positive real axis such that \( g(x) = \int_0^x t f(t) \, dt \). If \( g(x^3) = x^6 + x^7 \), then the value of \( \sum_{r=1}^{15} f(r^3) \) is:
- Let $f: [0, 3] \to A$ be defined by $f(x) = 2x^3 - 15x^2 + 36x + 7$ and $g: [0, \infty) \to B$ be defined by $g(x) = \frac{x}{x^{2025} + 1}$. If both functions are onto and $S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \}$, then $n(S)$ is equal to:
- Let A, B, C be three points in the xy-plane, whose position vectors are given by \[ \sqrt{3} \hat{i} + \hat{j}, \quad \hat{i} + \sqrt{3} \hat{j}, \quad \text{and} \quad a\hat{i} + (1-a) \hat{j} \] respectively with respect to the origin \( O \). If the distance of the point \( C \) from the line bisecting the angle between the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) is \( \frac{9}{\sqrt{2}} \), then the sum of all possible values of \( a \) is:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,