Parabola - 7
Dump Question: The
was written as Y and 
was
written as y.how?
Ans: The lines 3x-4y=0 and 4x+3y+12=0 are
to each other as clear from
their slopes. so, the rotation of axis and transformation of axes was performed
to get the desired easy form of parabola.
Q.2:- Three normals from a point to the parabola y2 = 4ax meet the axis of the parabola in points whose abscissa
are in A.P. Find the locus of the point. Ans : The equation of any normal to
the parabola is Y = mx - 2am -am3 It passes through the point (h,k) if am3
+ m (2a - h) + k = 0 ??????. (1) the normal cuts the axis of the parabola viz
, y = 0 at point where x = 2a + am2 hence the abscissa of the point in which
the normal through (h,k) meet the axis of the parabola are. X1 = 2a + am12 , x2 = 2a + am23 , x3 = 2a + am3 2 Since X1 ,x2, x3 are in A.P. (2a + am21) +( 2a + am23) = 2 (2a + am22)
= m12+m22=2m22..........(2)
also from (1) m1+m2+m3=0..............(3)
from (3) (m1+m2)2=m22==m12+m32+2m1m3=m22
Since m2 is a root of (1), am2 3+m2(2a-h)+k=0
= 2 k+m 2 (2a-h) + k = 0
27 ak2 = 2 (h - 2a)3
Hence the locus of(h,k) is 27ay2=2(x-2a)3 M-3- prove that the length of the intercept on the normal at the point p(at2,2at) of a parabola
y2=4ax made by the circle on the line joining the focus and point p as diameter is a
ANS: let the normal at p(at2,2at) cut the circle in k and the axis of parabola at g then pk is required intercept.
SP=PM=a+at2
Since angle in a semicircle being right angle .
SPR=90.
and normal at p(at2,2at) is
y = - tx+2at+at3
= tx + y - 2at - at3=0
SK is the perpendicular distance from s(a,0) to the normal(1)
in 
SPK
(PK)2 = (SP)2 - (SK)2
= a2(1+t2)2 - a2t2 (1 + t2)
= a2 (1+t2)
Dumb Question:- why is SP=PM in the above question? Ans: the point p lies on the parabola . so the distance of p from directrix is same as the distance from the focus. and hence SP=PM. Q-4:- Three normals are drawn from the point (0,0) to the currve y2=x. show that c must be greater than 1/2. one normal is always the x-axis . Find c for which the other two normals are perpendicular to each other.
Ans: Equation to normal to the parabola
is passing through (c,o)
Then only one normal will be there i.e , x axis
Now the slope of other two normal are
For these normals to be to each other
we need m2.m3=-1
so,
M-3:- Find the equation to the common tangents to the circle x2+y2=2a2
and the parabola y2=8ax.
Ans :-
The Equation of any tangent to the parabola y2=8ax i.e,
will touch the circle x2+y2=2a2 if
radius =
= length of the perpendicular from the centre(0,0) to the line
or, m2(1+m2)=2
or, m4+m2-2=0
or,(m2+2)(m2-1)=0
But m2+2=0 gives non real values of m.
m2-1=0;
m=
Putting m in (1) we get the equation of common tangents are
the equations of common tangents are.
y = x 2a
or y=x + 2a and y=-(x+2a)
Dumb question : why y= x2a gave only four equations of straight line
y=x+2a & y= -(x+2a) where as four lines are possible?
Ans:- Note that the equation of tangent in y=
So, it is not possible m with x be +ve and m with 2a be -ve . hence y=x-2a & y=-x+2a are impossible solutions.
Q.7: Prove that the circle circumscribing the triangle formed by tangents to a parabola passes through the focus
Ans: Let P(at12, 2at1),q(at22,2at2) and r(at32,2at3) be three points on the parabola y2=4ax
The equation of tangents at P't11 and Q't2' are
Y t11 = x + at12
Y t2 = x + at22
Solving these, x = at1t22 and y = a(t1 + t2 )
The point of intersection C of the tangents at P and Q is
{ at1 t2 , a(t1 + t2 )}.
similarly other points of intersection of tangents are
B = {at3t1, a(t3+t1)},A={at2t3,a(t2+t3)}
let the equation of circumcircle of the
ABC be
x2+y2+2gx+2fy+c=0------------(1)
(1) will pass through the focus (a,0) if
a2 + 2ga + c=0..........................(2)
A,B,C are points on (1) .so-
a2 t12 t22 +a2 (t1+t2)2+ 2g.at1t2+2.f.a(t1+t2)+c=0.........(3)
a2 t22 t32 +a2 (t2+t3)2+ 2g.at2t3+ 2.f.a(t2+t3)+c=0...........(4)
a2 t32 t12 +a2 (t3+t1)2+ 2g.at3t1+ 2.f.a(t3+t1)+c=0...........(5)
(3)-(4)
a2 t22( t12-t 32+a2 (t1-t3)(t1+2t2 +t3+2g.at2(t1- t3)+2.f.a(t1-t1)+c=0
or, a2 t22 (t1+t3+a2(t1+2t2 +t32g.at2t2+ 2.f.a=0..........(6)
similarly (4)-(5)
=> a2[t22 (t1 + t3) - 32(t2 + t1) + t2 - t3] + 2g.a(t2 - t3) = 0
or,a [t2t3(t2 - t3) + t1(t22 - t32) + t2-t3)] + 2g(t2-t3) = 0
or,a[t2t3 + (t1t2) + t1t3 + 1] + 2g=0
2g=-a(1 + t1t2 + (t2t3) + t3t1)
From t3x(6) - t2 X(7),we get
2f=-a(t1 + t2 + t3-t1t2t3)
putting values of 2g, 2f in (3) we get
the equation of the circumcircle is-
-
it passes through(a,0) because.-
-
Hence the problem.-
was written as Y and was written as y.how?
Ans: The lines 3x-4y=0 and 4x+3y+12=0 are
to each other as clear from their slopes. so, the rotation of axis and transformation of axes was performed
to get the desired easy form of parabola.
Q.2:- Three normals from a point to the parabola y2 = 4ax meet the axis of the parabola in points whose abscissa
are in A.P. Find the locus of the point. Ans : The equation of any normal to
the parabola is Y = mx - 2am -am3 It passes through the point (h,k) if am3
+ m (2a - h) + k = 0 ??????. (1) the normal cuts the axis of the parabola viz
, y = 0 at point where x = 2a + am2 hence the abscissa of the point in which
the normal through (h,k) meet the axis of the parabola are. X1 = 2a + am12 , x2 = 2a + am23 , x3 = 2a + am3 2 Since X1 ,x2, x3 are in A.P. (2a + am21) +( 2a + am23) = 2 (2a + am22)
= m12+m22=2m22..........(2)
also from (1) m1+m2+m3=0..............(3)
from (3) (m1+m2)2=m22==m12+m32+2m1m3=m22
Since m2 is a root of (1), am2 3+m2(2a-h)+k=0
= 2 k+m 2 (2a-h) + k = 0
27 ak2 = 2 (h - 2a)3
Hence the locus of(h,k) is 27ay2=2(x-2a)3 M-3- prove that the length of the intercept on the normal at the point p(at2,2at) of a parabola
y2=4ax made by the circle on the line joining the focus and point p as diameter is a
ANS: let the normal at p(at2,2at) cut the circle in k and the axis of parabola at g then pk is required intercept.
SP=PM=a+at2
Since angle in a semicircle being right angle .
SPR=90.and normal at p(at2,2at) is
y = - tx+2at+at3
= tx + y - 2at - at3=0
SK is the perpendicular distance from s(a,0) to the normal(1) in SPK(PK)2 = (SP)2 - (SK)2
= a2(1+t2)2 - a2t2 (1 + t2)
= a2 (1+t2)
Dumb Question:- why is SP=PM in the above question? Ans: the point p lies on the parabola . so the distance of p from directrix is same as the distance from the focus. and hence SP=PM. Q-4:- Three normals are drawn from the point (0,0) to the currve y2=x. show that c must be greater than 1/2. one normal is always the x-axis . Find c for which the other two normals are perpendicular to each other.
Ans: Equation to normal to the parabola
is passing through (c,o)Then only one normal will be there i.e , x axis
Now the slope of other two normal are
For these normals to be
to each otherwe need m2.m3=-1
so,
M-3:- Find the equation to the common tangents to the circle x2+y2=2a2
and the parabola y2=8ax.
Ans :-
The Equation of any tangent to the parabola y2=8ax i.e,
will touch the circle x2+y2=2a2 if
radius =
= length of the perpendicular from the centre(0,0) to the line
or, m2(1+m2)=2
or, m4+m2-2=0
or,(m2+2)(m2-1)=0
But m2+2=0 gives non real values of m.
m2-1=0;m=Putting m
in (1) we get the equation of common tangents are the equations of common tangents are.y =
x 2aor y=x + 2a and y=-(x+2a)
Dumb question : why y=
x2a gave only four equations of straight line y=x+2a & y= -(x+2a) where as four lines are possible?
Ans:- Note that the equation of tangent in y=
So, it is not possible m with x be +ve and m with 2a be -ve . hence y=x-2a & y=-x+2a are impossible solutions.
Q.7: Prove that the circle circumscribing the triangle formed by tangents to a parabola passes through the focus
Ans: Let P(at12, 2at1),q(at22,2at2) and r(at32,2at3) be three points on the parabola y2=4ax
The equation of tangents at P't11 and Q't2' are
Y t11 = x + at12
Y t2 = x + at22
Solving these, x = at1t22 and y = a(t1 + t2 )
The point of intersection C of the tangents at P and Q is
{ at1 t2 , a(t1 + t2 )}.
similarly other points of intersection of tangents are
B = {at3t1, a(t3+t1)},A={at2t3,a(t2+t3)}
let the equation of circumcircle of the
ABC bex2+y2+2gx+2fy+c=0------------(1)
(1) will pass through the focus (a,0) if
a2 + 2ga + c=0..........................(2)
A,B,C are points on (1) .so-
a2 t12 t22 +a2 (t1+t2)2+ 2g.at1t2+2.f.a(t1+t2)+c=0.........(3)
a2 t22 t32 +a2 (t2+t3)2+ 2g.at2t3+ 2.f.a(t2+t3)+c=0...........(4)
a2 t32 t12 +a2 (t3+t1)2+ 2g.at3t1+ 2.f.a(t3+t1)+c=0...........(5)
(3)-(4)
a2 t22( t12-t 32+a2 (t1-t3)(t1+2t2 +t3+2g.at2(t1- t3)+2.f.a(t1-t1)+c=0
or, a2 t22 (t1+t3+a2(t1+2t2 +t32g.at2t2+ 2.f.a=0..........(6)
similarly (4)-(5)
=> a2[t22 (t1 + t3) - 32(t2 + t1) + t2 - t3] + 2g.a(t2 - t3) = 0
or,a [t2t3(t2 - t3) + t1(t22 - t32) + t2-t3)] + 2g(t2-t3) = 0
or,a[t2t3 + (t1t2) + t1t3 + 1] + 2g=0
2g=-a(1 + t1t2 + (t2t3) + t3t1)From t3x(6) - t2 X(7),we get
2f=-a(t1 + t2 + t3-t1t2t3)
putting values of 2g, 2f in (3) we get
the equation of the circumcircle is--it passes through(a,0) because.-
-Hence the problem.-