Ans:- It was given in the question kthat it is to lbe shown that y = x - 3 is nirmal .
So, m = 1 should be factor of m3- 13m + 12 = 0 which it is.
Hence m3- 13 + 12 = 0 is written as
(m -1 ) (m2 + m - 12 ) = 0
or (m -1) (m + 4) (m - 3) = 0
Chord of contact .
Equation of chord of contact: -
T 0
or yy1 - 2a (x + x) = 0.
Why ?
Let point p(x1, y1) be
A (p1q ) and B (h1K ) on the parabola.
Equation of PB is lKy - 2a(x + h) = 0
Both these stangents pass through p(x + x1)
9y1 = 2a (x1 + p)
and Ky1= 2a (x 1 + h)
Now consiedr equation yy1 - 2a (x + x1 ) = 0
This equation is of first degree in x & y and it represcnts straight lline passing through
A (p1q ) and B (h1K )
Hence, equation yy1- 2a (x + x1) = 0
represents equation of line A B which is called chord of contact of point (x11y 1)
Illustration 8.
From point (- 1, 2) tangent line are drawn to parabola y2 = 4ax. find equation of hord of contact . Also find area of triangle formed by hoed of contac and the tangents.
Ans. Chord of lcontact of 0(-1,2) is
yy1 = 2a (x + x1)
or y = x - 1
Solving with parabola y = 2 4x we get the points
The distance of point 0(-1, 2)from line y = x -1 (Chord of contact) is
Chord of mid point ( x1,y1)
Note: Only one such chord is possible .
Why?
Any line through point (x1, y1) is
(y - y1) = m(x - x1 ----(1)
Now wehave to derermine value of m1 its slope.
If it meets parabolain P( at1, 2at1 )and
then its equation is
y(t1 + t2) - 2x - 2at1t2 = 0 ----(2)
So, by companing slopes in (1) and (2) we get
Since (x1,y1) is mid point of
Putting
On reamanging, we get
yy1 - 2a(x + x1) = y12 - 4ax1
or
Eauation of chord.
(i) Equation of chord joing (x1),y1) and (x2, y2) is
y(y1 + y2) = 4ax + y1y2
(ii)Equation of chard joing (at12, 2at2) and(at12, 2at2) is
y(t1 + t2) = 2 (x + at1 t2)
Note that both these results can be obtained by finding the middle point of the line joinign the two points and then using chord with mid point equation.
Illustration 9.
Show that locus of mid point of any focal chord is y2 = 2ax - 2a2
Ans Let the mid point be (h1K)
So, the equation of this kchord is
i.e yK - 2a (x+h) = K2 - 4ah.
Now this chord is focal chord, si it must pass through (9,0).
=> (9,0) must satisty
yk -2a (x+h) = K2 - 4aK
or . 0 - 2a(a + h) = k2 - 4ah
or k2 - 2ah + 2a2 = 0
Locus is y22ax + 2a2 = 0
or y2 = 2ax - 2a2.
Diamcter
Thelocus of middle point of a system of || chords of a parabola is called a diamcter .
If y = mx + c represents a system of | | chord of parabola y2 = 4ax, then likne y = 2a/m is equation of diamcter and this will meet parabola at
Why?
The chords y = mx + c where c varies infersed the parabola y2 = 4ax at point y1 and y1
Now, y' and y" are roots of equation
Which is obtained by solving y = mx + c
With y2= 4ax .
Equation is Let the middle point of the chords be (h,k)
0
9y1 = 2a (x1 + p)
distance of point 0(-1, 2)from line y = x -1 (Chord of contact) is 
then its equation is 
Let the middle point of the chords be (h,k)
