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

[Kenny52]

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

 

[MarkFL]

The family of lines passing through $(1,-2)$ is (using the point-slope formula):

\(\displaystyle y=m(x-1)-2\)

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

\(\displaystyle m(x-1)-2=x^2\)

Write in standard quadratic form:

\(\displaystyle x^2-mx+(m+2)=0\)

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

\(\displaystyle m^2-4(1)(m+2)=0\)

\(\displaystyle m^2-4m-8=0\)

Application of the quadratic formula yields:

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

And so the two lines are:

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

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]