The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(2,1)
The equation of the given curve is y2=4x.
Differentiating with respect to x, we have:
2y \(\frac{dy}{dx}\)=4=\(\frac{dy}{dx}\)=2y
Therefore, the slope of the tangent to the given curve at any point (x, y) is given by
\(\frac{dy}{dx}\)=\(\frac{2}{y}\)
The given line is y = x + 1 (which is of the form y = mx + c)
∴ The slope of the line = 1 The line y = x + 1 is tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve.
Thus, we must have:
\(\frac{2}{y}\)=1
y=2
Now, y=x+1=x=y-1=x=2-1=1
Hence, the line y = x + 1 is tangent to the given curve at the point (1, 2).
The correct answer is A.
If \( x = a(0 - \sin \theta) \), \( y = a(1 + \cos \theta) \), find \[ \frac{dy}{dx}. \]
Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
If f (x) = 3x2+15x+5, then the approximate value of f (3.02) is
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
m×n = -1