<p>i've a test in math tomrow in my country and it is, let's say the sat I math section test of my country. but i am having some uncertianities about the formulas of the tanget to the parabola</p>
<p>in one of my books it says that the formula of the tangent of the equation of a prablola of the form y^2=bx is - 2yy1=b(x+x1) while another one says that the equation is 2yy1=b(x-x1)</p>
<p>the same think about the parabola of the form x^2=by. one says that the equation is 2xx11=b(y+y1) and the other says that it is 2xx11=b(y+y1). can you tell me which one fo the books is right, but within this day because tomorrow i have the test. thank you</p>
<p>Take the 2nd form, where the parabola is by = x^2 .</p>
<p>The slope (dy/dx) in general is (dy/dx) = (2x) / b
At (x1, y1), the slope is (2x1) / b , and this is the slope of the tangent at (x1,y1).</p>
<p>The straight line tangent to this parabola at (x1,y1) has a slope of (2x1/b), and passes through the point (x1,y1).</p>
<p>The general straight line equation is y = (slope).x + intercept</p>
<p>Here, y = (2x1/b) .x + intercept
Plug in y=y1, x=x1 to get intercept = y1 - (2/b) . x1^2
Use this in the straight line equation to get
y = (2x1/b).x + (y1 - (2/b).x1^2)
which simplifies to</p>
<h2>(y-y1) = (2x1/b).(x-x1) or b(y-y1) = 2 x1 (x-x1)</h2>
<p>For the first form of the equation bx = y^2, I was about to say that you should get the same final form, just swapping 'x' and 'y' throughout i.e.
b(x-x1) = 2 y1 (y-y1)</p>
<p>However, note that for any x, you now have two possible values of y, for this parabola. You may want to work it out in full, just in case.</p>