Question:

The triangle formed by the tangent to the curve $f(x)=x^2+bx-b$ a t the point (1,1) and the coordinate axes, lies in the first quadrant. If its area is 2 sq units, then the value of b is

Updated On: Aug 15, 2022
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The Correct Option is C

Solution and Explanation

Let $y=f(x)=x^2+bx-b$ The equation of the tangent at P (1, 1) to the curve 2y =$2x^2+2bx-2b$ is $y+1 = 2x.1 + 6(x+1) - 26$ $\therefore$ $y = (2+b)x-(1+b)$ Its meet the coordinate axes at $x_A=\frac{1+b}{2+b}$ and $y_b=-(1-b)$ $\therefore$ Area of $\Delta$ OAB = $\frac{1}{2}OA \times OB$ $-\frac{1}{2}\times\frac{(1+b)^2}{(2+b)}=2$ $\Rightarrow$ $(1+b)^2+4(2+b)=0 \Rightarrow b^2+6b+9=0$ $\Rightarrow (b+3)^2=0 \Rightarrow b=-3$
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Questions Asked in JEE Advanced exam

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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