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

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I don"t know what the slope of the tangent line will be. Is it the negative reciprocal?



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.


$y=x^2-3x-5$ ,$dy/dx= 2x-3$ 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 $y+5=3(x-3)$


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 $\color{blue}{(3, -5)}$


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 $x=-\frac{b}{2a}$

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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