Solving Hyperbola Problem with F(-5,0) and D(5,0) | PF - PD = \pm6

[threetheoreom]

when there is enough block ... here is my problem.

a hyperbola having F(-5,0) and D(5,0) and difference of focal radii 6.

We can use |PF - PD|= 6 or PF - PF = $$\pm$$6.

I apply the distance formula ..eleminate radicals and simplify.

I get

16X^2 -9y^2=144

Divinding both sides by 144 I get

The hyperbolic equation

(x^2/9)-(y^2/16) = 1

analysing to graph ( herein lie the problems )

First determining the intercepts ..it is easy to see the intercepts on the x-axis are 3 and -3 ( verified by makeing y 0 in the above equ.) and that making x 0 we can aslo varify that there arn't any y intercepts (-y^2/16)

( that I understand)

The book goes on to say to determine the extent of the graph

making for y we get y=$$\pm$$4/3 [square root] of x^2 -9 and
x = $$\pm$$ 3/4 [squre root] of y^2 + 16

y is real if and only if |x| >= 3 thus no part of the graph lies between -3 and 3 ... in the second x is real for all y variables $$\Leftarrow$$ someone explain that please.

this part is killing me, they said notice that y = $$\pm$$4/3 [square root] x^2 - 9 = $$\pm$$4/3x [square root] 1-9/x^2 . $$\Leftarrow$$ I don't get this I'm thinking they factored it and i should be able to see that of course i tried ..but i can't see how . [ block]

they the go on to say that as the absolute value of x gets larger the square root of 1- 9/x^2 get closer to 1 and therefore |x| gets larger the graphapproaches the lines y $$\approx$$$$\pm$$ 4/3x. thus as
|x| becomes larger the graph approaches the lines y= 4/3x and -3/4x. :grumpy:



Thanks guys
Orson

p.s sorry for the bad tex

 

[HallsofIvy]

when there is enough block ... here is my problem.

a hyperbola having F(-5,0) and D(5,0) and difference of focal radii 6.

We can use |PF - PD|= 6 or PF - PF = $$\pm$$6.

I apply the distance formula ..eleminate radicals and simplify.

I get

16X^2 -9y^2=144

Divinding both sides by 144 I get

The hyperbolic equation

(x^2/9)-(y^2/16) = 1

analysing to graph ( herein lie the problems )

First determining the intercepts ..it is easy to see the intercepts on the x-axis are 3 and -3 ( verified by makeing y 0 in the above equ.) and that making x 0 we can aslo varify that there arn't any y intercepts (-y^2/16)

( that I understand)

The book goes on to say to determine the extent of the graph

making for y we get y=$$\pm$$4/3 [square root] of x^2 -9 and
x = $$\pm$$ 3/4 [squre root] of y^2 + 16

y is real if and only if |x| >= 3 thus no part of the graph lies between -3 and 3 ... in the second x is real for all y variables $$\Leftarrow$$ someone explain that please.

Solve for x: Adding $y^2/16$ to both sides, $x^2/9= 1+ y^2/16= (1/16)(16+ y^2)$. Now multiply both sides by 9 to get $x^2= (9/16)(16+ y^2)$. Finally, take the square root of both sides: $x= \pm (3/4)\sqrt{16+ y^2}$. Since a square is never negative, $16+ y^2$ is never less than 16 and so $\sqrt{16+ y^2}$ is never less than 4. Then x cannot be between -(3/4)(4)= -3 and (3/4)(4)= 3.

this part is killing me, they said notice that y = $$\pm$$4/3 [square root] x^2 - 9 = $$\pm$$4/3x [square root] 1-9/x^2 . $$\Leftarrow$$ I don't get this I'm thinking they factored it and i should be able to see that of course i tried ..but i can't see how . [ block]

No factoring involved. Solve for y, just as we solved for x above. Adding $y^2/16$ to both sides and subtracting 1 from both sides, we get $y^2/16= x^2/9- 1= (1/9)(x^2- 9)$. Multiply both sides of the equation by 16 to get $y^2= (16/9)(x^2- 9)$. Now take the square root of both sides to get $y= \pm (4/3)\sqrt{x^2- 9}$. Finally, factor that "x²" out of the square root. $(x^2- 9)= x^2(1- 9/x^2)$ and $\sqrt{x^2}= \pm x$. We have $y= \pm (4/3)x\sqrt{1- 9/x^2}$.

they the go on to say that as the absolute value of x gets larger the square root of 1- 9/x^2 get closer to 1 and therefore |x| gets larger the graphapproaches the lines y $$\approx$$$$\pm$$ 4/3x. thus as
|x| becomes larger the graph approaches the lines y= 4/3x and -3/4x. :grumpy:

No need to be grumpy! Yes, for very large x, $9/x^2$ will be very close to 9 (and getting closer as x gets larger) so $\sqrt{1- 9/x^2}$ will be very close to 1 (and getting closer as x gets larger and larger). That means that $y= \pm (4/3)x\sqrt{1- 9/x^2}$ will be very close to $y= \pm (4/3)x$. The two straight lines, y= (4/3)x and y= (-4/3)x are the "asymptotes" of this hyperbola- the hyperbola gets closer and closer to those straight lines as x gots to plus or minus infinity.

A good way of sketching this hyperbola is: Mark the vertices (-3,0) and (3,0) on the graph. Draw the rectangle with horizontal sides from (-3,4) to (3,4), (-3,-4) to (3,-4) and vertical sides from (-3,-4) to (-3, 4), (3,-4) to (3, 4). How draw the diagonals of that rectangle, extending them as far as you please outside the rectanle. Those are the asymptotes y= (4/3)x and y= (-4/3)x. Finally, starting from (3, 0), draw a smooth curve upward and to the right getting closer and closer to the asymptote y= (4/3)x. Draw a smooth curve starting at (3,0) downward and to the right getting closer and closer to the asymptote y= (-4/3)x. That give you the right half of the hyperbola. Starting from (-3,0) draw a smooth curve upward and to the left getting closer and closer to the asymptote y= (-4/3)x and a smooth curve from (-3,0) to the left and downward getting closer and closer to the asymptote y= (4/3)x. That gives you the left half of the hyperbola.

Thanks guys
Orson

p.s sorry for the bad tex

Suggestion- use "itex" rather than "tex" for things you want in a line with other text. Keep "tex" for formulas on a separate line. Also, put whole formulas in tex, not just individual symbols.

 

[threetheoreom]

Thanks hallofIvy great explanation. Yeah the computations were getting caught in the explanation n the book and in my head, that also interseting method for sketching .. i can also plug some appropriate values for x in in the equation for y.

Thanks again