MHB How to Find Lines Tangent to a Parabola Passing Through a Given Point?

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To find the lines tangent to the parabola y = x² that pass through the point (1, -2), the equation of the lines can be expressed as y = m(x - 1) - 2. Substituting this into the parabola's equation leads to a quadratic equation in standard form: x² - mx + (m + 2) = 0. For the line to be tangent, the discriminant must be zero, resulting in the equation m² - 4m - 8 = 0. Solving this using the quadratic formula gives two slopes, m = 2(1 ± √3), leading to the equations of the two tangent lines.
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Find the equation of the straight line(s) which pass through the point (1, −2) and is (are) tangent to the parabola with equation y = x2

No calculus is to be used.I can substitute the point into the equation for the straight line giving -2=m+c

And into the parabola (-2)2 = m+c

Not sure if this is getting me anywhere
 
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The family of lines passing through $(1,-2)$ is (using the point-slope formula):

$$y=m(x-1)-2$$

Now, we may substitute for $y$ in $y=x^2$ to obtain:

$$m(x-1)-2=x^2$$

Write in standard quadratic form:

$$x^2-mx+(m+2)=0$$

Since the line is tangent to the parabola, we require the discriminant to be zero:

$$m^2-4(1)(m+2)=0$$

$$m^2-4m-8=0$$

Application of the quadratic formula yields:

$$m=2\left(1\pm\sqrt{3}\right)$$

And so the two lines are:

$$y=2\left(1\pm\sqrt{3}\right)(x-1)-2$$

Here's a graph:

[DESMOS=-10,10,-10,10]y=x^2;y=2\left(1+\sqrt{3}\right)\left(x-1\right)-2;y=2\left(1-\sqrt{3}\right)\left(x-1\right)-2[/DESMOS]
 
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