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Chapter 15: Problem 1406

The line \(\mathrm{y}=\mathrm{mx}+1\) is a tangent to the parabola \(\mathrm{y}^{2}=4 \mathrm{x}\) if \(\mathrm{m}=\ldots \ldots \ldots \ldots\) (a) 4 (b) 3 (c) 2 (d) 1

Short Answer

Expert verified
The line \(y= mx+1\) is tangent to the parabola \(y^2 = 4x\) if \(m=1\), which corresponds to choice (d).

Step by step solution

01

Replace y in the parabola equation with the line equation

Using the line equation \(y = mx+1\), replace y in the parabola equation \( y^2 = 4x \): \((mx+1)^2 = 4x\)
02

Expand and simplify the equation

Expanding and simplifying, we get: \(m^2x^2 + 2mx + 1 = 4x\) This equation is a quadratic equation in terms of x. If the line is tangent to the parabola, the discriminant of the quadratic equation should be equal to 0.
03

Find the discriminant and create the discriminant equation

By comparing the given equation with the general form \(ax^2 + bx + c = 0\), we can identify: \( a = m^2, b = 2m - 4, c = 1 \) The discriminant of the quadratic equation, D, is given by: \( D = b^2 - 4ac \) So we have, \(D = (2m-4)^2 - 4(m^2)(1)\) Since we want the discriminant to be equal to 0: \((2m-4)^2 - 4(m^2) = 0\)
04

Solve for the value of m

Now, solve this discriminant equation for m: \((2m-4)^2 - 4(m^2) = 4m^2 - 16m + 16 - 4m^2 = 0\) Simplifying and solving for m: \(-16m + 16 = 0 \Rightarrow m = 1\) Thus, the value of m is 1, which corresponds to the choice (d).

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