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I'm stuck with problem supposed to be trivial.
I need to find tangent line witch touches curve $y^2 = -4ax$ at the point $(x_0,y_0)$

Rewriting it as $$x = -\frac {y^2}{4a}$$ Taking derivative: $$f(y)'=-\frac {y}{2a}$$ Then slope is $$f(y_0)'=-\frac {y_0}{2a}$$ Then make a substitution in tangent line equation at the point $(x_0,y_0)$ to find the offset:
$$x = ky+b \iff b = x-ky \iff b = x_0 -\frac {y_0^2}{2a}$$ Then write our tangent line equation: $$x = -\frac {y_0y}{2a} + x_0 -\frac {y_0^2}{2a} \iff 2ax = -y_0y + 2ax_0 -y_0^2$$

But this result doesn't match with result given in the book: $$4ax+2yy_0-y_0^2 = 0$$ Help me out, guys.

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You computed $\frac{dx}{dy}$ at $(x_0,y_0)$. This is the reciprocal of the slope. Easy fix, the slope at $(x_0,y_0)$ is $-\frac{2a}{y_0}$.

Alternately, differentiate implicitly. We get $$2y\frac{dy}{dx}=-4a.$$ So the slope of the tangent line at $(x_0,y_0)$ is $-\frac{4a}{2y_0}$.

Remark: Your procedure should also have worked. However, there is a mistake in the next to last displayed line. It should be $2ax_0$, not $2x_0$. (When you cleared denominators by multiplying through by $2a$, the term $x_0$ did not get multiplied by $a$.)

Edit: The error has been corrected. To get the answer in the book, note that $2ax_0=-\frac{y_0^2}{2}$.

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  • $\begingroup$ Thank you for the answer. But, for starters, i don't see why I have to flip my derivative. As it obvious from my question, I look at equation like y is a dependant variable (Yes, it's non-conventional, but can't take square root here and solve for x). So, why would I flip then? And even if I flip, I get $x = - \frac {2ay} {y_0} + x_0 -2a$ $\endgroup$ – wf34 May 10 '14 at 16:47
  • $\begingroup$ Agree with your remark. I lost an a. Already made the edit, thanks. $\endgroup$ – wf34 May 10 '14 at 16:49
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas May 10 '14 at 16:50
  • $\begingroup$ What about my previous comment? Your suggestion gives result that also doesn't match to result given in the book. Honestly I don't see why it would be valid to do it your way. Please, review my previous comment. $\endgroup$ – wf34 May 10 '14 at 16:52
  • $\begingroup$ Your answer is now correct. To get the book's form, as my edit at the end suggests, use the fact that $2ax_0=y_0^2/2$. $\endgroup$ – André Nicolas May 10 '14 at 16:54
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There are a couple of mistakes in your working out (I see the second one has now been corrected in your question)

1. Where you write

$$b = x-ky \iff b = x_0 -\frac {y_0^2}{2a}$$

you should have instead

$$b = x-ky \iff b = x_0 +\frac {y_0^2}{2a} $$

2. In addition you forgot a factor of $a$ on the $x_0$ so

$$x = -\frac {y_0y}{2a} + x_0 +\frac {y_0^2}{2a} \iff 2ax = -y_0y + 2x_0 +y_0^2$$

becomes

$$x = -\frac {y_0y}{2a} + x_0 +\frac {y_0^2}{2a} \iff 2ax = -y_0y + 2ax_0 +y_0^2$$

Finally, you can just substitute for $2ax_0$ using $y_0^2=-4ax_0$, giving

$$ 2ax = -y_0y - \frac{y_0^2}{2} +y_0^2 $$

from which the result follows

$$ 4ax + 2y_0y - y_0^2 = 0 $$

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