* These equations represent mirror lines
to their plus counterparts. Both lines are parallel.
* In fact, x/b ± y/a =1 and y = mx ± c1
are various forms of the equation of a st. line. Which form one
chooses is a matter of convention. We have chosen the most used
format. * The equation ax+by+c=0 represents a
st.line whose x-intercept is b, y-intercept is a and c represents
the negative of the products of the intercepts i.e. c=-ab. Hence
ay+bx+c=0 represents a st. line whose intercepts have been
exchanged. *The equation of a straight line which
is drawn perpendicular to the line given by equation ax+by+c=0 and
passes through the point of interception(0,-c/b) or (0,a) of the
line with y-axis is given by b2x-aby-ac=0
or b2x-aby+a2b=0.
*The equation of a straight line which is drawn perpendicular to the
line given by equation ax+by+c=0 and passes through the point of
interception(-c/a,0) or (b,0) of the line with x-axis is given by
a2y-abx-bc=0 or or a2y-abx+ab2=0. * the perpendicular distance from a point
(x1,y1) to a st.line with equation ax + by + c =0
is given by d= (ax1+by1+c) / √ (a2 + b2 )
*In A general equation of the type
Ax2 +Bxy+Cy2 +Dx +Ey+F=0,
the eccentricity e is given by e = √(2 √[ (A-C)2 + B2 ])
/√ (n(A+C) +√[ (A-C)2 + B2 ] )
& n =1 if (AC-B2/4) < 0 and n = -1 if (AC-B2/4) > 0
Pl. note that D,E,F are not contributing to eccentricity. If B2
=4AC, curve is a parabola. The co-efficient B contributes to
eccentricity and solely contributes to the slanting of the
curve with respect to the co-ordinate axis. Since B2
is positive, its effect is to increase the eccentricity e whereas
the role of second factor AC depends on overall sign. Taking A'=2A
and C'=2C. we can write B2
= > <A'C' or B= > <√(A'C'). Let us state that M=√(A'C')= geometric
mean of A',C'. Hence B may be equal to, less than or greater than
the geometric mean. In case of non-equality, one can calculate the
standard deviation.
* In Ax2 +Bxy+Cy2 +Dx +Ey+F=0
if B2-4AC < 0 ellipse
if B2-4AC < 0 , A=C, B=0 , circle
if B2-4AC = 0 , parabola
if B2-4AC > 0 , hyperbola
if B2-4AC > 0, A+C=0, Rectangular hyperbola
^ a1x1 +b1y1+c1=0 ; co-ordinate
of y-intercept (0,a1). length from this point to (x1.y1) is (x1-0)2
+(y1-a1)2 = L2
Let the st.line perpendicular to the above line, having a point
(x2,y2) which is analogous to (x1,y1) have the equation
a2x2+b2y2+c2=0 .......(II)
and length from this point to (x2.y2) to (0,a1) is
(x2-0)2
+(y2-a1)2 = n1=L2 ......(I)
Squaring & rearranging both sides of .....(II), we get
a22x22-b22y22-c22-2b2c2y2=0
.......(III)
multiplying both sides of (I) by
a22 , we get
a22x22+a22y22-2a22a1y2+a22a21-n2=0
.......(IV) where n2=a22 *n1
(IV) - (III) = y22(a22+b22)
+2y2(b2c2-a22a1 ) +(c22+a22a21
-n2)=0
Taking A= (a22+b22),
B= 2(b2c2-a22a1 ), C=(c22+a22a21
-n2)
this is a quadratic equn A y22
+By2 +C=0
y2=-B ± √( B2 -4AC) /2A
x2= (-C2-b2y2)/a2 ;
when the rotation is through the x-intercept point, the equn becomes
(IVa) - (IIIa) = x22(a22+b22)
+2x2(a2c2-b22b1 ) +(c22+b22b21
-n'2)=0
where
n'2=b22 *n'1
and n'= (x2-b1)2
+(y2)2 = L2
and
A= (a22+b22),
B= 2(a2c2-b22b1 ), C=(c22+b22b21
-n'2)
this is a quadratic equn A x22
+Bx2 +C=0
x2=-B ± √(B2-4AC) / 2A
y2= (-C2-a2x2)/b2 ;
|