In Exercises 5–12, find the standard form of the equation of each...

Question

In Exercises 5–12, find the standard form of the equation of each hyperbola satisfying the given conditions. Foci: (−4, 0), (4, 0); vertices:(−3, 0), (3, 0)

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where we want to find the standard form equation of a hyper below. With the conditions that has focus at negative five and zero five and zero and vortices at negative 10 and positive 10. So in order to find the equation for this hyperbole, let's remember what the standard form equation is. So in this case be why values for all of the coordinates are staying the same. So that tells me that is transverse across the X axis. So the standard form equation for that is x squared divided by a squared minus y squared divided by b squared is equal to one. And from this standard form you can see that we need to determine the values of A. And B. So for that let's also remember that the coordinates for the folk I are given to us in terms of h plus or minus C and K. And the coordinates for the vertex is are given to us in terms of h plus or minus A and Kay and the center is equal to H K. So from The coordinates we can also see that the folk I go from negative 5 to positive five which means that the midpoint or the center between the folk I is zero. So our center is actually at 00 or the origin Which means that H is equal to zero and K. is equal to zero. So that helps us simplify these other coordinates that were given. So the folk I that were given is -5 and zero. And we can compare that to h minus C and K. Where we already knew that K is equal to zero. But now we have a church minus C is equal to negative five. And we know H is zero. So this becomes negative, C Is equal to -5. If we divide by a negative we get C is equal to five and we can solve the same way for our Vertex. So say I use one and 0. We can compare that to H plus A. K. And again we already knew that K is equal to zero. But now we have a church plus A. Is equal to one. And because H is zero, I have A is equal to one. So notice I've sold for A and for C but I don't have a value for B. So that is where our third we're not third but our equation to relate A B and C come or I have C squared is equal to a squared plus B squared. And if I solve for B I'll subtract a squared from both sides. I'll have B squared is equal to c squared minus a squared. If I take the square root of both sides, I have B is equal to the square root of c squared minus a squared. And I know the values of C. N. A. So I can plug those in. So B is equal to the square root of c squared which is five squared minus a squared which is one. If I simplify that square root I have B is equal to the square root of 25 minus one, or B is equal to The Square Root of 24. So now I finally have a value for B. And I can plug this back into my standard form equation. So I have X squared divided by a squared which is one minus Y squared divided by B squared Which is the square root of 24 squared is equal to one. And if I simplify this I have x squared divided by one minus Y squared divided by the square root of 24 times. The square root of 24 is just 24. It's equal to one. And notice I don't need to write the divided by one. So I can write this as X squared minus y squared, divided by 24 is equal to one. And this becomes my final standard form equation for this hyperbole to and you can see that that aligns with answer choice D. So hopefully this video helped and see you next time

Struggling with College Algebra?

In Exercises 5–12, find the standard form of the equation of each hyperbola satisfying the given conditions. Foci: (−4, 0), (4, 0); vertices:(−3, 0), (3, 0)

Watch next

Hyperbolas at the Origin practice set