The slope of the tangent to a curve C : y=y(x) at any point [x, y) on it is $\frac{2 e ^{2 x }-6 e ^{- x }+9}{2+9 e ^{-2 x }}$ If C passes through the points $\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right) $ and $\left(\alpha, \frac{1}{2} e ^{2 \alpha}\right)$ $ then $e ^\alpha$ is equal to :

$\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

The slope of the tangent to a curve C : y=y(x) at any point [x, y) on it is $$\frac{2 e ^{2 x }-6 e ^{- x }+9}{2+9 e ^{-2 x }}$$ If C passes through the points $$\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right) $$ and $$\left(\alpha, \frac{1}{2} e ^{2 \alpha}\right)$$ $ then $$e ^\alpha$$ is equal to :}

$\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

$\frac{3+\sqrt{2}}{3-\sqrt{2}}$

$\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

$\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$

$\frac{\sqrt{2}+1}{\sqrt{2}-1}$

The slope of the tangent to a curve C : y=y(x) at any point [x, y) on it is 2e^2x-6e^-x+9/2-^2x +9e If C passes through the points then $e ^\alpha$ is equal to :

Questions Asked in JEE Main exam

View More Questions

Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points.

A straight line can be represented as an equation in various forms, as show in the image below:

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y₀ = m (x – x₀)

2. Two – Point Form

Let's look at the line. L crosses between two places. P₁(x₁, y₁) and P₂(x₂, y₂) are general points on L, while P (x, y) is a general point on L. As a result, the three points P₁, P₂, and P are collinear, and it becomes

The slope of P₂P = The slope of P₁P₂, i.e.

$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$

Hence, the equation becomes:

y - y₁ =$ \frac{y_2-y_1}{x_2-x_1} (x-x1)$

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c

The Correct Option is B

Solution and Explanation

Top Questions on Straight lines

Questions Asked in JEE Main exam

Concepts Used:

Straight lines

1. Slope – Point Form

2. Two – Point Form

3. Slope-Intercept Form