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 + am32 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), am23+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
was written as Y and 
was 
to each other as clear from 
SPR=90.
SK is the perpendicular distance from s(a,0) to the normal(1)
SPK
is passing through (c,o)
ABC be
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