In geometry, a tangent line to a curve is defined as any straight line that touches the curve at only one point. The equation of a tangent parabola can be found by finding the intersection of two lines through (2, −3) and using the standard form for an equation in slope-intercept form.

- find equations of both lines through the point (0, −14) that are tangent to the parabola y = x² + x
- find equation of line through (-11,-13) and (15,-17).
- identify how many points on a curve it takes for two straight lines not intersecting one another can still be considered as being parallel.
- find equation of line through (0, −14) and (−11,-13).
- find equation of a tangent to the curve y = x² + x.
- find three equations for lines passing through point (-15, 15) that are not parallel but intersect at one point.

Find Equations of Both Lines Through Point (0, -14) That Are Tangent to Parabola y = x²+x: The first step in finding an equation for a parabola is determining its intercepts, which can be found by solving two equations with different coordinates on the parabola’s axes. In this case, the intercepts for the parabola y = x²+x are (0, -14) and (−11,-13). Because these two points have different coordinates, they give rise to a linear equation that is parallel to both of them.

## Find Equation of Line Through (15,-17): The line passing through point (-15, 15) has an equation given by:

y=\frac{|-x^22} {(-|x)} + \frac{|-x^23}{(-|x)} + \frac{\cos q }{45} where | represent absolute value signs. This says that if you take every possible combination of values for x from −16 up until 16 including 0 and find the corresponding value for y, you will find that all of them satisfy this equation. This is because there are two values

for x on both sides of it’s coefficient in front of cos q, which means we can say “y=x+q” when solving this linear equation. However, since our desired point (15,-17) has coordinates (-15,-14), substituting those into the original equation results in: \frac{|-x^22} {(-|x)} + \frac{|-x^23}{(-|x)} = \frac{\cos(-√25 )} {45}, or equivalently √25 with a negative sign equals −√25 } + \frac{|-x^23}{(-|x)} + \frac{\cos q }{45} where | represent absolute value signs. This says that if you take every possible combination of values for x from −16 up until 16 including 0 and find the corresponding value for y, you will find that all of them satisfy this equation. This is because there are two values for x on both sides of it’s coefficient in front of cos q, which means we can say “y=x+q” when solving this linear equation. However, since our desired point (15,-17) has coordinates (-15,-14), substituting those into the original equation results in: \frac{|-x^24}{(-|x)} + \frac{\cos q }{45} = \frac{\cos(-√25 )} {45}, which is not satisfied by any combination of values for x. This means that the point does not solve this equation, so it cannot be tangent to the parabola.

### The first thing we need to do is find what our equation should look like if | represent absolute value signs instead of negative numbers (our desired point has coordinates (−15,-14), substituting these into an original equation would yield:

(\frac{|-x^27}{(-|x)}\)\). Since there are two different solutions when solving a linear equation for both x and y, we need to find the equation that has two different solutions. Taking a look at our original equation: \frac{|-x^24}{(-|x)} = \frac{\cos q }{45}, we can see that this is in fact one of those equations with two different solutions (think about what happens when you solve the first quadrant). This means that there are an infinite number of lines passing through point (-15,-14) tangent to the parabola; however, only one will be satisfying since it’s impossible for more than one line to pass through a single point. So now all we have to do is find which solution satisfies \(\frac{|-x^24}{(-|x)+|-y^15} = \frac{\cos q }{45}\). and find the equation that has two different solutions. Taking a look at our original equation: \frac{|-x^27}{(-|x)} = \frac{\cos q }{45}, we can see that this is in fact one of those equations with two different solutions (think about what happens when you solve the first quadrant). This means that there are an infinite number of lines passing through point (-15,-14) tangent to the parabola; however, only one will be satisfying since it’s impossible for more than one line to pass through a single point.

So now all we have to do is find which solution satisfies \(\frac{|-x^24}{(-|x)+|-y^15} = \frac{\cos q }{45}\). The answer is the one in which x=−14. Plugging this into our original equation yields: \( \left(\ -\ 14, -0.39)\) and \( (0, 0)\), respectively. Doing the same for y leads us to an infinite number of solutions since it will depend on how you plug in a value for x; however,

### only one will be satisfying since it’s impossible for more than one line to pass through a single point. So there are two tangent parabolas: \(\left(\ -\ 14, 0.39)\) and \( (0, y)\).

This is a tangent parabola because the two lines are both passing through point (14,-14), which means that they’re also going to be intersecting at some point. Of course we can’t find this intersection without knowing what value of x goes into the equation; so let’s try plugging in −x=−15 for all values of x=-∞≤x≤+∞ and see where our equations go wrong. Now we need to solve the following system of differential equations: \begin{align*} dx & = |-y^24| \\ dy & = -x^34 \end{align*}

If x=-∞, then the first equation is undefined. This means that our system will not be solvable for any values of x below this value because it makes the second differential equation \(dy = -x^34\) undefined as well. If we let xt=+∞ instead, then both equations are satisfied and two lines tangent to the parabola y = x² + x are found: \(\left(\ -\ 14, 0.39)\) and \( (0, y)\). For all other values of xt≠-∞ or −∞ there would only be one line passing through point (14,-14), which can’t be tangent to the parabola.