Question:

A straight line parallel to the line 2x - y + 5 = 0 is also a tangent to the curve $y^{2}=4x+5.$ Then the point of contact is

Updated On: Apr 26, 2024
  • (2,1)
  • (-1,1)
  • (1,3)
  • (3,4)
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The Correct Option is B

Solution and Explanation

Given curve is $y^{2}=4x+5.$ on differentiating, we get $2 y \frac{d y}{d x}=4 \, \Rightarrow\quad\frac{d y}{d x}=\frac{2}{y}$ Given line is 2x - y + 5 = 0 $\Rightarrow\, y=2x+5$ slope of line is 2. Therefore, $\frac{2}{y}=2 \Rightarrow y=1$ put y = 1 in the equation of curve, we get 1 = 4x + 5 x = - 1 Hence, point of contact is (- 1, 1)
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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