The slope of normal at any point \((x, y), x > 0, y > 0\) on the curve \(y = y(x)\) is given by \(\frac{x^2}{xy-x^2y2^-1}\) If the curve passes through the point (1, 1), then \(e · y(e)\) is equal to
- \(\frac{1-tan(1)}{1+tan(1)}\)
- \(tan(1)\)
- \(1\)
- \(\frac{1+tan(1)}{1-tan(1)}\)
The Correct Option is D
Solution and Explanation
\(∵ -\frac{ dx}{dy} = \frac{x^2}{xy-x^2y^2-1}\)
\(\frac{dy}{dx} = \frac{x^2y^2-xy+1}{x^2}\)
Assuming \(xy = v ⇒ y+x \frac{dy}{dx} = \frac{dv}{dx}\)
\(\frac{dv}{dx}-y = \frac{(v^2+v+1)y}{v}\)
\(\frac{dv}{dx} =\frac{v^2+1}{x}\)
\(∵ y(1) = 1 ⇒ tan^{–1} (xy) = lnx + tan^{–1}(1)\)
Put \(x\) = \(e\) and \(y\) = \(y(e)\) we get
\(tan^{–1} (e · y(e)) = 1 + tan^{–1} 1\).
\(tan^{–1} (e · y(e)) – tan^{–1} 1 = 1\)
\(∴ e(y(e)) = \frac{1+tan(1)}{1-tan(1)}\)
Hence, the correct option is (D): \(\frac{1+tan(1)}{1-tan(1)}\)
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Concepts Used:
The Slope of a Line
A slope of a line is the conversion in y coordinate w.r.t. the conversion in x coordinate.
The net change in the y-coordinate is demonstrated by Δy and the net change in the x-coordinate is demonstrated by Δx.
Hence, the change in y-coordinate w.r.t. the change in x-coordinate is given by,
\(m = \frac{\text{change in y}}{\text{change in x}} = \frac{Δy}{Δx}\)
Where, “m” is the slope of a line.
The slope of the line can also be shown by
\(tan θ = \frac{Δy}{Δx}\)
Read More: Slope Formula
The slope of a Line Equation:
The equation for the slope of a line and the points are known to be a point-slope form of the equation of a straight line is given by:
\(y-y_1=m(x-x_1)\)
As long as the slope-intercept form the equation of the line is given by:
\(y = mx + b\)
Where, b is the y-intercept.