find specific points of a tangent line on a curve.

Q: Find the coordinates of all points on the graph of $\displaystyle y=1-x^2$ at which the tangent line passes through the point $\displaystyle (2,0)$

__My solution so far:__

What I understood is that there is a point on outside the parabola/ curve $\displaystyle (2,0)$ and another point is on the curve which is unknown and need to figure out $\displaystyle (x, y)$ and they are connected through a tangent line.

Now first let me figure out the derivative/slope of the tangent line over that curve.

$\displaystyle y=1-x^2$

$\displaystyle f(x)=1-x^2$ { is y always equals f(x)? maybe this is only when y is function of x right?}

$\displaystyle f'(x)=-2x$

So now we know the slope of the tangent line

we know 2 co-ordinates of one end $\displaystyle (2,0)$ and need to figure out the other end $\displaystyle (x,y)$.

and we can use the slope formula here

$\displaystyle \frac{0-y}{2-x}=-2x$

$\displaystyle -y=-2x(2-x)$

$\displaystyle -y=-4x+2x^2$

$\displaystyle -(1-x^2)=-4x+2x^2$ [as $\displaystyle 1-x^2=y$ from the original equation]

$\displaystyle 1-x^2=4x-2x^2$

$\displaystyle 0=4x-2x^2+x^2-1$

$\displaystyle 0=4x-x^2-1$

$\displaystyle 0=-x^2+4x-1$

No easy way to factor this so I used the formula

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

So that makes it

$\displaystyle x=\frac{-4\pm\sqrt{4^2-4(-1)(-1)}}{2(-1)}$

$\displaystyle x=\frac{-4\pm\sqrt{16-4}}{-2}$

$\displaystyle x=2\pm\sqrt{12}$

$\displaystyle x=2\pm2\sqrt{3}$

so basically we got 2 xs

$\displaystyle x=2+2\sqrt{3}$ and $\displaystyle x=2-2\sqrt{3}$

where did I do wrong..

why the answer says $\displaystyle 2\pm\sqrt{3}$ not sure if that is x or y.. looks close to x...

Re: find specific points of a tangent line on a curve.

Quote:

Originally Posted by

**ameerulislam** Q: Find the coordinates of all points on the graph of $\displaystyle y=1-x^2$ at which the tangent line passes through the point $\displaystyle (2,0)$

__My solution so far:__

What I understood is that there is a point on outside the parabola/ curve $\displaystyle (2,0)$ and another point is on the curve which is unknown and need to figure out $\displaystyle (x, y)$ and they are connected through a tangent line.

Now first let me figure out the derivative/slope of the tangent line over that curve.

$\displaystyle y=1-x^2$

$\displaystyle f(x)=1-x^2$ { is y always equals f(x)? maybe this is only when y is function of x right?}

$\displaystyle f'(x)=-2x$

So now we know the slope of the tangent line

we know 2 co-ordinates of one end $\displaystyle (2,0)$ and need to figure out the other end $\displaystyle (x,y)$.

and we can use the slope formula here

$\displaystyle \frac{0-y}{2-x}=-2x$

$\displaystyle x=2\pm2\sqrt{3}$ **CORRECT**

**BUT** $\displaystyle x=2\pm2\sqrt{3}\ne \pm 4\sqrt{3}$

$\displaystyle (2\pm2\sqrt{3})^2=4\pm4\sqrt{3}+12$

Re: find specific points of a tangent line on a curve.

Quote:

Originally Posted by

**Plato** **BUT** $\displaystyle x=2\pm2\sqrt{3}\ne \pm 4\sqrt{3}$

$\displaystyle (2\pm2\sqrt{3})^2=4\pm4\sqrt{3}+12$

yeah yeah i figured that our immediately and edited., but why the bok is saying $\displaystyle x=2\pm\sqrt{3}$ 2 is missing. so they are wrong and I didn't make any other mistakes?

Re: find specific points of a tangent line on a curve.

ok I think I found it..

that basic mistake was done up there as well.