Find a point on the curve y = (x − 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Solution and Explanation
If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord.
The slope of the chord is \(\frac{4-0}{4-2}\) =\(\frac42\)=2.
Now, the slope of the tangent to the given curve at a point (x, y) is given by,
\(\frac{dy}{dx}\)=2(x-2)
Since the slope of the tangent = slope of the chord, we have:
2(x-2) = 2
x-2=1=x=3
when x=3,y=(3-2)2=1
Hence, the required point is (3, 1)
Top Questions on Applications of Derivatives
- Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f''(x)>0 \) for all \( x \in \mathbb{R} \) and \( f'(a-1) = 0 \), where \( a \) is a real number. Let \( g(x) = f(\tan^2 x - 2\tan x + a) \), \( 0<x<\frac{\pi}{2} \).
Consider the following two statements :
(I) \( g \) is increasing in \( (0, \frac{\pi}{4}) \)
(II) \( g \) is decreasing in \( (\frac{\pi}{4}, \frac{\pi}{2}) \)
Then,- JEE Main - 2026
- Mathematics
- Applications of Derivatives
- If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.
- CBSE CLASS XII - 2026
- Mathematics
- Applications of Derivatives
- Find the absolute maximum value of $f(x)=\cos x+\sin^2 x$, where $x\in[0,\pi]$.
- CBSE CLASS XII - 2026
- Mathematics
- Applications of Derivatives
- Find the equation of the tangent to the curve \(y = x^2\) at the point (1,1).
- BITSAT - 2025
- Mathematics
- Applications of Derivatives
- A boat takes 4 hours to travel 40 km downstream and 6 hours to return. The speed of the stream is:
- VITEEE - 2025
- Mathematics
- Applications of Derivatives
Questions Asked in CBSE CLASS XII exam
- The rate for the following reaction is given by: \[ A + B \rightarrow C, \quad \text{Rate} = k[A][B]^2 \] How is the rate affected if we double the concentration of B?
- CBSE CLASS XII - 2026
- Chemical Kinetics
- Find the general solution of the differential equation \[ y\log y\,\frac{dx}{dy}+x=\frac{2}{y}. \]
- CBSE CLASS XII - 2026
- Differential Equations
- A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:
- CBSE CLASS XII - 2026
- Electromagnetic induction
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following:
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
- CBSE CLASS XII - 2026
- Integration
- Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]
- CBSE CLASS XII - 2026
- Current Density
Concepts Used:
Tangents and Normals
- A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
- Given a function y = f(x), the equation of the tangent for this curve at x = x0
- Slope of tangent (at x=x0) m=dy/dx||x=x0
- A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
- The normal to a given curve y = f(x) at a point x = x0
- The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
Diagram Explaining Tangents and Normal:
