(5) Transverse axis is y-axis and hyperbola is centralised (centre at (0,0) so foci are on y-axis at (0,c) and (0,-c) where c is focal length. Equation form is:
y2/a2-x2/b2=1 and c2=a2+b2, c2/a2=1+b2/a2. The latus recta are horizontal (in this case) chords passing through each of the foci. e (eccentricity)=c/a=√(1+b2/a2)=2√3 (given), and c2/a2=12.
When y=±c, then c2/a2-x2/b2=1, x2=b2(c2/a2-1)=11b2, x=±b√11. Therefore the end points of a latus rectum are (b√11,c) and (-b√11,c), making the length=2b√11=18 (given), b√11=9, b=9/√11; b2=81/11.
1+b2/a2=12=1+(81/11)/a2,
11a2=81/11,
a2=81/121.
[c2=a2+b2=81/121+81/11=81(132/121), c=18√3/11, c/a=2√3.]
The equation of the hyperbola is 121y2/81-11x2/81=1 or 121y2-11x2=81.
(6)
(a) y2=12x is a "sideways" parabola.
(b) 9x2-16y2+36x+32y+124=0 can be written:
9(x2+4x+4-4)-16(y2-2y+1-1)+124=0.
9(x+2)2-16(y-1)2-36+16+124=0,
9(x+2)2-16(y-1)2+104=0,
16(y-1)2/104-9(x+2)2/104=1, which is a hyperbola with transverse axis being vertical (x=-2), centre at (-2,1).
(c) x2+9y2=9 is an ellipse centre at the origin with semi-major axis (x) = length 3 and semi-minor axis (y) = length 1.
Rewrite (d): x2-10x+25-25+y2-4y+4-4-21=0,
(x-5)2+(y-2)2-25-4-21=0,
(x-5)2+(y-2)2=50 is a circle centre (5,2), radius 5√2.
(e) Should be: 4y2-12xy+10x2+2x+1=0. When x=0: 4y2+1=0, so y cannot be represented graphically. When y=0, 10x2+2x+1=0, which has no solution, so x cannot be represented graphically. If the constant term had been -1, it would have been a tilted ellipse. But all this shows is that the curve intersects none of the axes. If the figure exists then some shift of the axes will allow such intersections. So let X=x-h and Y=y-k, therefore x=X+h and y=Y+k. Substitute in the equation:
4(Y+k)2-12(X+h)(Y+k)+10(X+h)2+2(X+h)+1=0,
4Y2+8kY+4k2-12XY-12kX-12hY-12hk+10X2+20hX+10h2+2X+2h+1=0. Let constant term K=4k2-12hk+10h2+2h+1:
4Y2+8kY-12XY-12kX-12hY+10X2+20hX+2X+K=0. Let X=0:
4Y2+(8k-12h)Y+K=0, from the quadratic formula, Y=(3h-2k±√((3h-2k)2-K))/2, so (3h-2k)2-K≥0, (3h-2k)2≥K, and, substituting for K:
9h2-12hk+4k2≥4k2-12hk+10h2+2h+1, which simplifies to:
0≥h2+2h+1, 0≥(h+1)2, which has the only solution h=-1. Therefore, since X=0, x=X+h=-1, and K=4k2+12k+9=(2k+3)2.
(3h-2k)2-K=(3h-2k)2-(2k+3)2=(3h-2k-2k-3)(3h-2k+2k+3)=3(3h-4k-3)(h+1)=0 when h=-1 already, so we don't have to consider the factor 3h-4k+3.
So, since Y=y-k, (3h-2k)/2=y-k, 3h-2k=2y-2k, y=3h/2. And we already have x=h=-1, so y=-3/2.
The equation therefore represents the singularity point (-1,-3/2). So it has no shape. No other values of x and y satisfy the given equation.
(f) 8x2-4xy+5y2-144=0 is a tilted ellipse.
Express this in polar coordinates: x=rcosθ, y=rsinθ:
8r2cos2θ-12r2sinθcosθ+5r2sin2θ=144.
Any point P(r,θ) on the graph is represented by a line OP from the origin inclined by θ to the x-axis.
If the x-axis is now tilted by angle φ, anticlockwise, OP is now inclined by an angle θ-φ, but its length r is unaffected. Let's suppose the new axes are X and Y. We need to express the polar equation of the graph in terms of X and Y. The new coordinates of P with respect to the new axes are:
X=rcos(θ-φ) and Y=rsin(θ-φ), which can be expanded using trig identities:
X=rcosθcosφ+rsinθsinφ, Y=rsinθcosφ-rcosθsinφ.
We can substitute x for rcosθ and y for rsinθ:
X=xcosφ+ysinφ, Y=ycosφ-xsinφ. We can find x and y from this system of equations:
Xcosφ=xcos2φ+ysinφcosφ, Ysinφ=ysinφcosφ-xsin2φ.
Therefore: Xcosφ-Ysinφ=x.
Xsinφ=xsinφcosφ+ysin2φ, Ycosφ=ycos2φ-xsinφcosφ.
Therefore: Xsinφ+Ycosφ=y.
Now we can substitute x and y in the equation 8x2-4xy+5y2-144=0:
8(Xcosφ-Ysinφ)2-4(Xcosφ-Ysinφ)(Xsinφ+Ycosφ)+5(Xsinφ+Ycosφ)2=
8X2cos2φ-16XYsinφcosφ+8Y2sin2φ
-4X2sinφcosφ-4XYcos2φ+4XYsin2φ+4Y2sinφcosφ
+5X2sin2φ+10XYsinφcosφ+5Y2cos2φ-144=0.
We want all the XY terms to cancel out so that we can get the equation of the rectified ellipse (i.e., not tilted).
Therefore:
-16sinφcosφ-4cos2φ+4sin2φ+10sinφcosφ=0, which simplifies to:
2sin2φ-3sinφcosφ-2cos2φ=0, now divide through by cos2φ:
2tan2φ-3tanφ-2=0=(2tanφ+1)(tanφ-2), φ=tan-1(-½) or tan-1(2) (these are perpendicular gradients, representing perpendicular axes).
The equations of the X and Y axes are y=-x/2 and y=2x. sinφ=-√5/5, sin2φ=⅕, cosφ=2√5/5, cos2φ=⅘, sinφcosφ=⅖. The negative value simply means clockwise rotation rather than anticlockwise rotation.
So 8X2cos2φ-4X2sinφcosφ+5X2sin2φ+8Y2sin2φ+4Y2sinφcosφ+5Y2cos2φ=144,
(8cos2φ-4sinφcosφ+5sin2φ)X2/144+(8sin2φ+4sinφcosφ+5cos2φ)Y2/144=1 is the form of an ellipse. The trig quadratics cannot be negative so this ensures that we have an ellipse and not some other figure.
a and b (the "radii" of the ellipse) can be found having found φ.
The coefficients evaluate to (32/5+8/5+1)/144=1/16 and (8/5-8/5+4)/144=1/36, making a=6 (semi-major) and b=4 (semi-minor), making the fundamental equation of the ellipse X2/36+Y2/16=1 or X2/16+Y2/36, depending on how the rotation is viewed.