Question:
The area of the figure bounded by the parabola \((y-2)^2 = x-1\), the tangent to it at the point with the ordinate 3 and the \(x\)-axis is
The area of the figure bounded by the parabola \((y-2)^2 = x-1\), the tangent to it at the point with the ordinate 3 and the \(x\)-axis is
Show Hint
Integrate with respect to \(y\) since parabola is sideways.
Updated On: Apr 23, 2026
- 3
- 6
- 9
- None of these
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ (y-2)^2 = x-1 \Rightarrow x = (y-2)^2 + 1 \]
Step 2: Calculation / Simplification}
At \(y = 3\): \(x = (3-2)^2 + 1 = 2\). Point: \((2, 3)\)
\(\frac{dx}{dy} = 2(y-2)\). At \(y=3\): \(\frac{dx}{dy} = 2\)
Tangent: \(x - 2 = 2(y - 3) \Rightarrow x = 2y - 4\)
Intersection with \(x\)-axis: \(y=0 \Rightarrow x = -4\)
Area = \(\int_0^3 [((y-2)^2 + 1) - (2y-4)] dy\)
\(= \int_0^3 (y^2 - 4y + 4 + 1 - 2y + 4) dy = \int_0^3 (y^2 - 6y + 9) dy\)
\(= \left[\frac{y^3}{3} - 3y^2 + 9y\right]_0^3 = 9 - 27 + 27 = 9\)
Step 3: Final Answer
\[ 9 \]
\[ (y-2)^2 = x-1 \Rightarrow x = (y-2)^2 + 1 \]
Step 2: Calculation / Simplification}
At \(y = 3\): \(x = (3-2)^2 + 1 = 2\). Point: \((2, 3)\)
\(\frac{dx}{dy} = 2(y-2)\). At \(y=3\): \(\frac{dx}{dy} = 2\)
Tangent: \(x - 2 = 2(y - 3) \Rightarrow x = 2y - 4\)
Intersection with \(x\)-axis: \(y=0 \Rightarrow x = -4\)
Area = \(\int_0^3 [((y-2)^2 + 1) - (2y-4)] dy\)
\(= \int_0^3 (y^2 - 4y + 4 + 1 - 2y + 4) dy = \int_0^3 (y^2 - 6y + 9) dy\)
\(= \left[\frac{y^3}{3} - 3y^2 + 9y\right]_0^3 = 9 - 27 + 27 = 9\)
Step 3: Final Answer
\[ 9 \]
Was this answer helpful?
0
0
Top MET Mathematics Questions
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- The equation $3^{3x + 4} = 9^{2x - 2},\ x>0$ has the solution
- A parallelogram is constructed on the vectors $\mathbf{a} = 3\alpha - \beta$, $\mathbf{b} = \alpha + 3\beta$. If $|\alpha| = |\beta| = 2$ and the angle between $\alpha$ and $\beta$ is $\dfrac{\pi}{3}$, then the length of a diagonal of the parallelogram is
- Every group of order 7 is
- Let $U_{n} = 2 + 2^{3} + 2^{5} + \cdots + 2^{2n+1}$ and $V_{n} = 1 + 4 + 4^{2} + \cdots + 4^{n-1}$. Then $\displaystyle\lim_{n \to \infty} \dfrac{U_n}{V_n}$ is equal to
View More Questions
Top MET limits and derivatives Questions
- The value of \( \lim_{n \to \infty} \frac{1 + 2 + 3 + \dots + n}{n^{2}} \) is:
- \( \lim_{x \to \infty} \left(\frac{x+5}{x+2}\right)^{x+3} \) equals
- The value of \( \lim_{x \to 0} \left(\frac{e^x + 2^x + 4^x}{3}\right)^{\frac{2}{x}} \) is:
- If \( f(x) = \begin{cases} \frac{\sin |x|}{x}, & \text{for } [x] \ne 0 \\ 0, & \text{for } [x] = 0 \end{cases} \) where, \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\lim_{x \to 0} f(x)\) is equal to
- If \[ \lim_{x \to 0} \frac{\sin(\sin x) - \sin x}{ax^3 + bx^5 + c} = -\frac{1}{12}, \] then
View More Questions
Top MET Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
View More Questions