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At what point of the parabola $y=x^2-3x-5$ is the tangent line parallel to $3x-y=2$? Find its equation.

I don't know what the slope of the tangent line will be. Is it the negative reciprocal?

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5 Answers 5

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To be parallel, two lines must have the same slope.

The slope of the tangent line at a point of the parabola is given by the derivative of $y= x^2-3x-5$.

This means that the question is asking at what point the derivative of the parabola will equal the slope of $3x-y=2$.

So, to solve the problem, identify the slope of the line and set it equal to the derivative of the equation of the parabola to find the $x$ value of the point you want. Then use the equation of the parabola to find the $y$ value, and you're done.

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  • $\begingroup$ P.S. the slope of a line given by $y=mx+b$ is $m$. $\endgroup$
    – coldnumber
    Commented Jul 23, 2015 at 5:11
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$y=x^2-3x-5$ ,$dy/dx= 2x-3$ slope of the tangent line parallel to $3x-y=2$ ,whose slope is 3 which means $ 3=2x-3$ i.e. $x=3 $,$y=-5$ and the equation of tangent will be $y+5=3(x-3)$

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Edit: since the tangent is parallel to the given line: $3x-y=2$ hence the slope of tangent line to the parabola is $\frac{-3}{-1}=3$

Let the equation of the tangent be $y=3x+c$

Now, solving the equation of the tangent line: $y=3x+c$ & the parabola: $y=x^2-3x-5$ by substituting $y=3x+c$ as follows $$3x+c=x^2-3x-5$$ $$\implies x^2-6x-(c+5)=0\tag 1$$ For tangency we have the following condition $$ \text{determinant},\ B^2-4AC=0$$ $$\implies (-6)^2-4(1)(-(c+5))$$ $$\implies c=\frac{-56}{4}=-14$$ Hence, setting the value of $c=-14$ we get $$x^2-6x-(-14+5)=0$$ $$\implies x^2-6x+9=0$$ $$\implies (x-3)^2=0\implies x=3$$ Now, setting the value of $x=3$ in the equation of parabola as follows $$y=(3)^2-3(3)-5=-5$$ Hence, the point of tangency is $\color{blue}{(3, -5)}$

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  • $\begingroup$ The question was not asking for the intersection of the parabola and the line. $\endgroup$
    – coldnumber
    Commented Jul 23, 2015 at 5:35
  • $\begingroup$ @coldnumber: You are absolutely right, I misunderstood the statement. $\endgroup$ Commented Jul 23, 2015 at 5:58
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If you solve simultaneously the curve and the line $y=3x+c$ to get a quadratic equation in $x$ then this quadratic must have double roots at the point of tangency. This will give the value of $c$ and the required $x$ value is given by $x=-\frac{b}{2a}$

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Slope of tangent to parabola $y=x^2-3x-5$ is parallel to slope of line $3x-y=2$ when the angle between them is $0$. Given parabola $$y=x^2-3x-5$$ Differentiating w.r.t. "x" $$\frac{dy}{dx}=2x-3$$ Slope of tangent to parabola $$m_1=\frac{dy}{dx}=2x-3$$ Given line $$3x-y=2$$ Just to write in the form of $y=mx+c$ $$y=3x-2$$ So the slope of line is $$m_2=3$$ Now to find the angle between two slope we have $$tan\theta = \frac{m_2-m_1}{1+m_2m_1}$$ $$tan\theta = \frac{3-2x+3}{1+3(2x-3)}$$ Here we need to find the value of "$x$"

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