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Find equation of a line with gradient equal to $2$ which intersects the parabola $y = 6 − x − x^2$ at one point?

I tried using the equation of the line $y=2x+c$ and making it equal to $y= -x^2 - x +6$. Then finding $x$ using the quadratic formula. But I just can't find the $c$ value.

Micah
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Dzoker
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  • I tried using equation of line y=2x+c and make it equal to y= -x^2 - x +6. Then find x using quadratic formula. But I just can't find the c value. – Dzoker Jul 25 '15 at 11:46
  • Hint: let the determinant of the quadratic equation $+x^2+3x+c-6=0$ be equal to zero. Then there will be only one point of intersection, and will get the value of c. – Cataline Jul 25 '15 at 11:57

2 Answers2

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Let's find point of intersection: $$ 2x+c=-x^2-x+6\Longrightarrow x^2+3x+(c-6)=0. $$ These equation must have exactly one root; so, $$ 3^2=4\cdot1\cdot(c-6)\Longrightarrow c=\frac{33}{4} $$

Michael Galuza
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The equation of such a line, which passes through the point $(x_0,y_0)$ is given by $$y-2x=y_0-2x_0$$ It intersects the parabola at only one point if the abscissae equation for the intersection (which is a quadratic equation): $$6-x-x^2=2x+y_0-2x_0\iff x^2+3x+y_0-2x_0-6=0$$ has a double root, which amounts to saying its discriminant is $0$: $$\Delta=0\iff9=4\cdot(y_0-2x_0-6)\iff y_0-2x_0=\frac{33}4$$ Thus an equation of the line with slope $2$, tangent to the parabola, is $$y-2x=\frac{33}4\enspace\text{or}\quad 8x-4y+33=0.$$ We then can compute the abscissa of the point of contact, solving $$x^2+3x+\frac{33}4-6\iff4x^2+12x+9=(2x+3)^2=0\iff x=-\frac32$$ whence $y=-3+\dfrac{33}4=\dfrac{21}4$.

Bernard
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