# Thread: Tangent to a parabola through a given point.

1. ## Tangent to a parabola through a given point.

Hi guys

Thanks

2. Well, set up your favorite form of an equation of a line. It should probably have an x, a y, and two or three more variables (parameters) in it. The fact that the line passes through (2,-3) should give you an equation and/or some other fact about the parameters. The fact that the line is tangent to that parabola should give you more information.

Alternatively, find the tangent line to the parabola at a general point, and see which of those lines pass through (2,-3).

3. I tried to solve the problem and I got the equation for the first line y=-x-1 which i guess is right, but i could not find the equation of the second line... can you help me

4. Originally Posted by metallica007
I tried to solve the problem and I got the equation for the first line y=-x-1 which i guess is right, but i could not find the equation of the second line... can you help me
Equation of tangent at any point on the given parabolla is given by

$\frac{y + y_1}{2} = xx_1 + \frac{x + x_1}{2}$

This tangent passes through (2, -3). Put x = 2 and y = -3 in the above equation and solve for y1. Put this value in the equation of the aprabolla and solve for x1.

You get two values for x1. Find the corresponding values for y1.

Substitute these values in the equation of tangent.

5. Originally Posted by sa-ri-ga-ma
Equation of tangent at any point on the given parabolla is given by

$\frac{y - y_1}{2} = xx_1 + \frac{x + x_1}{2}$

This tangent passes through (2, -3). Put x = 2 and y = -3 in the above equation and solve for y1. Put this value in the equation of the aprabolla and solve for x1.

You get two values for x1. Find the corresponding values for y1.

Substitute these values in the equation of tangent.
this is the first time i knew about the equetion $\frac{y - y_1}{2} = xx_1 + \frac{x + x_1}{2}$ !!
guys i am completely lost, can you please solve the problem for me please ..

6. Originally Posted by metallica007
this is the first time i knew about the equetion $\frac{y - y_1}{2} = xx_1 + \frac{x + x_1}{2}$ !!
guys i am completely lost, can you please solve the problem for me please ..
Sorry. There is typo.

It should be (y+y1)/2.

7. The given parabola is $y= x^2+ x$ which has derivative $2x_0$ when x= x_0. The slope of any tangent to that parabola must be $2x_0$ for some number $x_0$.

Also, any line through (2, 3) is of the form y- 3= m(x- 2) so a line through (2, 3) tangent to the parabola is of the form $y- 3= (2x_0)(x- 2)$.

Of course, the tangent line must touch the parabola at $x= x_0$, that is when $x= x_0$, $y= x_0^2+ x_0$

That is, you must have $x_0^2+ x_0- 3= 2x_0(x_0- 2)$. Solve that for $x_0$ and then write the equation of the tangent lines. (That is a parabola so there will be two solutions.)

8. Originally Posted by metallica007
Hi guys

Thanks
Here comes a slightly different approach to this problem:

1. A line passing through (2, -3) has the equation:

$y-(-3)=m(x-2)~\implies~y=mx-2m-3$

3. Calculate the coordinates of the points of intersection of the parabola and the line:

$x^2+x=mx-2m-3~\implies~x^2+(1-m)x+2m+3=0$

Solve for x:

$x = \dfrac{-(1-m) \pm \sqrt{(1-m)^2-4 \cdot 1 \cdot (2m+3)}}{2}$

$x = \dfrac{m-1}2 \pm \dfrac12 \cdot \sqrt{m^2-10m-11}$

3. A line is tangent to the parabola if there exist only one point of intersection: the tangent point. This is only possible if the dicriminant equals zero:

$m^2-10m-11=0~\implies~m=-1~\vee~m=11$

Plug in these values into the equation of the line at 2.

4. The x-coordinate of the tangent point is $\dfrac{m-1}2$. Plug in the values of m to get the x-coordinates and plug in the x-coordinates into the equation of the line to get the corresponding y-coordinates of the tangent points.

5. For confirmation draw a sketch of the parabola and the two lines. (My sketch isn't drawn to scale!)