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?
[Math] Tangent line parallel to another line
calculusderivatives
Related Solutions
Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.
The tangent at $P$ is the angle bisector of $\angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.
But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.
Rewrite the equation of the ellipse $x^2+4y^2=8$ as $$ \begin{align} x^2+4y^2&=8\\ 4y^2&=8-x^2\\ y^2&=\frac{8-x^2}{4}\\ y&=\left(\frac{8-x^2}{4}\right)^{\frac12}.\tag1 \end{align} $$
The slope of the lines that are tangent to the ellipse is $$ \begin{align} \frac{dy}{dx}&=\frac{d}{dx}\left(\frac{8-x^2}{4}\right)^{\frac12}\\ &=\frac12\left(\frac{8-x^2}{4}\right)^{-\frac12}\left(\frac{-2x}{4}\right)\\ &=\frac12\left(\frac{4}{8-x^2}\right)^{\frac12}\left(-\frac{1}{2}x\right)\\ &=-\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12} \end{align} $$ or the quickest way $$ \frac d{dx}(x^2+4y^2=8)\quad\Rightarrow\quad 2x+8y \frac {dy}{dx}=0\quad\Rightarrow\quad \frac {dy}{dx}=-\frac {x}{4y}. $$ Since it is parallel with the line $x +2y = 6$, then $$ \begin{align} -\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12}&=-\frac{1}{2}\\ x\left(\frac{1}{8-x^2}\right)^{\frac12}&=1.\tag2 \end{align} $$ Square both sides of $(2)$, yield $$ \begin{align} \frac{x^2}{8-x^2}&=1\\ x^2&=8-x^2\\ 2x^2&=8\\ x_{1,2}&=\pm2\tag3 \end{align} $$ $y$ can be found by plugging in $(3)$ to $(1)$, yield $$ \begin{align} y_{1,2}&=\left(\frac{8-2^2}{4}\right)^{\frac12}\\ &=\pm1, \end{align} $$ where $(x_{1,2},y_{1,2})$ are the points of contact. Now, the equation of tangent line can be found by using $$ \frac{y-y_{1,2}}{x-x_{1,2}}=-\frac12. $$
Best Answer
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.