Dumb Question . Why two aprabolas are possible ?

Ans In this question only lasus ractum is given and two parabola are possible having same latus ractum on either side as shown in figure

Tangents to a parabola :
Let us find equation of tangents to parabola y² = 4ax

(i)Equation of tangent of slope m : -
y = mx + a/m
And, the point of contact is p = (a/m², 2a/m)

Why ?
Let line y = mx + c . be tangent to parabola y ² = 4ax.
Saluing the two curres tom gether.
We get, (mx + c)² = 4ax .
Since the line is tangent, this equation will have repeated roots.
So, D = O.
Now, (mx)² + (2mc - 4a) x + c² = 0
is equation.
D = 4 (mc-2a)² - 4m² c ² = 0
=, (mc - 2a)² - m²c² = 0
=, (mc - 2a - mc)(mc - 2a + mc) = 0
=| - 4a (mc - a) = 0
=, c = a/m

(ii) Equation of tangent at (x₁,y₁)
yy₁ = 2a (x+x₁)

Why?
Now y² = 4ax .
so, dy/dx = 2a/y
= is slope of tangent at (x₁,y ₁)
So,
Put m - 2a / y₁ in equation y = mx + a/m
=> y = 2a/y₁ x + ay₁/2a
=> yy₁ = 2ax + 4ax₁/2


(iii) Equation of tangent at (at ², 2at)
ty = x + at²

Why ?
Put x₁ = at ²| y₁ = 2at in
Equation yy₁ = 2a (x + x1)
=> (2at) ly = 2a (x + at)²
= ty = x + at²

Illustation 4.
Prove that line x + my + am² = 0 touches the parabola y ² = 4ax. Also find or dinates of point of contact .

Ans:- Solving, parabola y² = 4ax with line
x + my + am² = 0, we get
y² 4a (-my - am²)
or. y ² + 4amy + 4a²m² = 0
or. (y + 2am)² = 0 .
Since above is a perfect square, therefore both the values of y are equal. Hence given line is tangent to parabola and ordinates of the point to contact is - 2am (from y - 2am = 0)
Ptting, value of m in equation of the line we get x = am².
Hence point of contact is (am², - 2am)

Illustration 5
Find equation of straught lines touching both x²+ y² = 2a² and y² = 8a

Ans:- the parabola is y² = 8ax
or y² = 4 (2a) x .
So, equation of tangent is
y = mx + 2a/m
or. m²x - lmy + 2a = 0
Since this is tan gent to x² + y² = 2a², So the length of perpendicular from (0,0) must be equal to the radius ie

or
=> m⁴ + m² - 2 = 0
=> (m²+2)(m²-1) = 0
=> m²-1 = 0
=>
So, the equired tangent is


Equation of pair of tangents.
The equation of pair of tangent from (x₁,y ₁) is
t² = ss₁
or. (yy₁- 2a(x + x₁))² = (y² - 4ax)(y₁² - 4ax¹)

Why ?
Let p(x₁1 y₁) be point from which tangents is drawin to y² = 4ax


Let m(h₁R) be any point on either of tangent of straight line joining p and M is

As this is tangent to given parabola, it should be of the form.
y = mx + a/m .-------(3)
Comparing (2) and (3) we get

Eliminating M from equation(4) and (5) we get .
=> a(h - x₁)² = (K-y₁)(hy₁- K x₁)
Putting (x₁y₁) in place of (h₁K) gibes equation of locus of M. as
a(x - x₁)₂ = (y - y₁)(xy₁ - yx₁)
Which on rearranging gives .


Illustration 6.
Find the equation of tangents drawn to y² + 12x = 0from point (3,8).

Ans Now, y² = - 12x
or 4a = - 12
=> a = - 3
On . Comparing with standard form of tangent to parabola should be
y = mx + a/m or y = mx -3/m
Since tangent passes through (3,8) we ha
8 = 3m -3/m
or. 3mm² - 8m -3 = 0
or (m - 3) (3m+1) = 0
=> m = 3, -1/3
Hence there _L two tangents tnrough point (3,8)which are
3x - y -1 = 0
& x + 3y -27 = 0

Dumb Question Why tangent was taken in the form y = mx + a/m .?

Ans Whenever we have to take tangent from a point lying outside the parabola it is preferabee to take equation in this form as it is left second by satistying given opoint in the equation.

Normal to a parabola.
(1) Equation of normal at point (t)
y = -tx +2at +at³



Why?
Slope of normal at (at², 2at)
Equation of normal at (at²,2at) is y - 2at = - t (x-at²)
or y = - tx + 2at + at³

(2) Equation of normal slope m
Putting m = - t in above equation
y = mx - 2am - am³ (3) Equation of normal at (x₁,y₁)
y-y₁= -y₁/2a (x - x₁)

Some important point .
(1) Maximum 3 normals can be drawn from anyn piont (h₁K ) to parabola y² = 4ax .



Why?
Equation of normal is
y = - tx + 2at + at³
So, 3 different value of t are possible when we put (h₁ K) in given eqquation
i.r K = - t(n) + 2at + at³
Hence, 3different normals are possing .
(2) Three distirnt normals can be drawn from point (h₁ K ) to parabola y₂4qx iff 4 > 2a .

Illustration 7.
Three normal to y² = 4x pass through point (15,12) Show that one the normal is given by y =x - -3 and find equation of the others.

Ans:- y² = 4x .
=> 4a =4 ie a = 1.
Any normal to parabola is
y = mx - 2am - am³
putting a= 1 , we get
y = mx - 2am - am³
As it passses through (15,12) we get
12 = 15m - 2m - m³
or m³ - 13m + 12 = 0
or (m - 1) (m + 4) (m - 3) = 0
m = 1, 3; - 4.
Hence three naormals are
y = x - 3
y = -4x + 72
y = 3x - 33.