algebra-precalculus
If a line is tangent to one point and normal to another point on the curve $x=4t^2+3$,$y=8t^3-1$ then the question is to find out the slope of such a line.
Slope of curve at $(4t_1^2+3,8t_1^3-1)$ is $3t_1$.Now it is normal to the curve at point $t_2$,then slope of normal at that point is $\frac{-1}{3t_2}$.Equating we get $9t_1t_2=-1$.I couldn't proceed after this.Any help shall be highly appreciated. Thanks
You know that the tangent at the curve is given by
$$y_t = f(a)+f'(a)(x-a)$$
The normal would be a line such that
Given that a line that passes through $(X,Y)$ and has slope $m$ is given by
$$y-Y = m(x-X)$$
...and that two lines of slopes $n$ and $p$ are perpendicular if and only if $m\cdot n=-1$
Can you find $y_n$?
Given that a line that passes through $(X,Y)$ and has slope $m$ is given by
$$y-Y = m(x-X)$$
Give the equations to
See the sketch. We know $OT = 8, OP = 12$. So can you find $PT$ using Pythagoras?
Now using the slope of line through $AP$, you should notice that
$\displaystyle \frac{OT}{PT} = \frac{AO}{OP}~$
You know $OT, PT$ and $OP$. So can you find $AO$? Now the same length is obstructed below x-axis.
Best Answer
The equation of the tangent at the point $(4t^2+3,8t^3-1)$ is $$y-8t^3+1=3t(x-4t^2-3)$$
The tangent meets the curve again at $P(4p^2+3,8p^3-1)$ so substitute these coordinates into the tangent equation and we get, after cancellation, $$8p^3-8t^3=3t(4p^2-4t^2)$$
This can be simplified and factorised easily since we know that, due to tangency, $(p-t)^2$ must be a factor, and the only other solution is $$p=-\frac t2$$
The tangent is also the normal at $P$ so $$3p=-\frac{1}{3t}$$
Eliminating $p$ gives $$t=\pm\frac{\sqrt{2}}{3}$$ so the possible gradients are $$\pm\sqrt{2}$$