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Problem :

An equation of a tangent to the parabola $y^2=8x$ is $y=x+2$. the point on this line from which the other tangent to the parabola is perpendicular to the given tangent is given by ...

Solution :

Slope at any point on the parabola from where tangent can be drawn can be taken by differentiating equation of parabola $y^2=8x$

Which gives $2y \frac{dy}{dx}=8 \Rightarrow \frac{dy}{dx}=4/y$

Slope of the given line is equal to the slope of the line at this point therefore, $y =4$ and $x=2$.

Please suggest whether this is the correct method of doing this and how can I proceed this problem further thanks..

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By observing the function, we can see that it's a parabola opened to the right. A good thing to do would be to plot out the function to see what's going on. You can see the plot here: http://www.wolframalpha.com/input/?i=y%5E2+%3D+8x

One problem is that $y$ isn't a function of $x$. For the same $x$ value there are often two corresponding $y$ values. This will complicate things. However, we see that $x$ is a function of $y$. For each value of $y$, there is exactly one value of $x$. So an easy trick to deal with the problem is to exchange the $x$s and $y$s. So instead, solve the following problem:

An equation of a tangent to the parabola $x^2=8y$ is $x=y+2$. the point on this line from which the other tangent to the parabola is perpendicular to the given tangent is given by ...

Then, when you've solved it, exchange the $x$s and $y$s again. Plot the points on the original graph and make sure they jive with your intuition.

Best of luck!

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Hints:

  • You know the gradient of the tangent, so you can find the slope of a line perpendicular to it.

  • You can use the same method you have already used to find the point where this second line is tangent to the parabola and so the equation of the second tangent.

  • You now have the equation for two straight lines, and find where they intersect.

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$1)$ $y^2=8x$

$\frac{dy}{dx} = 8x$

 8---------(1

$\frac{dy}{dx}= y(2)$

$2(2)= 4$

$y-2=4(x-8)$ $y-2= 4x-32$

$y= 4x-16$ Tangent of the line.

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  • $\begingroup$ Please typeset your responses. $\endgroup$ – Alekos Robotis Feb 6 '16 at 7:08
  • $\begingroup$ This site uses MathJax formatting $\endgroup$ – JKnecht Feb 6 '16 at 7:18
  • $\begingroup$ Please explain in greater detail what you are doing in your answers. $\endgroup$ – JKnecht Feb 6 '16 at 7:19

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