# Parabola - 3

__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 y

^{2}= 4ax

(i)

__Equation of tangent of slope m__: -

y = mx + a/m

And, the point of contact is p = (a/m

^{2}, 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 + c

^{2}= 0

is equation.

D = 4 (mc-2a)

^{2}- 4m

^{2}c

^{2}= 0

=, (mc - 2a)

^{2}- m

^{2}c

^{2}= 0

=, (mc - 2a - mc)(mc - 2a + mc) = 0

=| - 4a (mc - a) = 0

=, c = a/m

(ii) Equation of tangent at (x

_{1},y

_{1})

yy

_{1}= 2a (x+x

_{1})

__Why__?

Now y

^{2}= 4ax .

so, dy/dx = 2a/y

= is slope of tangent at (x

_{1},y

_{1})

So,

Put m - 2a / y

_{1}in equation y = mx + a/m

=> y = 2a/y

_{1}x + ay

_{1}/2a

=> yy

_{1}= 2ax + 4ax

_{1}/2

(iii) Equation of tangent at (at

^{2}, 2at)

ty = x + at

^{2}

__Why__?

Put x

_{1}= at

^{2}| y

_{1}= 2at in

Equation yy

_{1}= 2a (x + x1)

=> (2at) ly = 2a (x + at)

^{2}

= ty = x + at

^{2}

__Illustation 4__.

Prove that line x + my + am

^{2}= 0 touches the parabola y

^{2}= 4ax. Also find or dinates of point of contact .

__Ans__:- Solving, parabola y

^{2}= 4ax with line

x + my + am

^{2}= 0, we get

y

^{2}4a (-my - am

^{2})

or. y

^{2}+ 4amy + 4a

^{2}m

^{2}= 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 = am

^{2}.

Hence point of contact is (am

^{2}, - 2am)

__Illustration 5__

Find equation of straught lines touching both x

^{2}+ y

^{2}= 2a

^{2}and y

^{2}= 8a

__Ans__:- the parabola is y

^{2}= 8ax

or y

^{2}= 4 (2a) x .

So, equation of tangent is

y = mx + 2a/m

or. m

^{2}x - lmy + 2a = 0

Since this is tan gent to x

^{2}+ y

^{2}= 2a

^{2}, So the length of perpendicular from (0,0) must be equal to the radius ie

or

=> m

^{4}+ m

^{2}- 2 = 0

=> (m

^{2}+2)(m

^{2}-1) = 0

=> m

^{2}-1 = 0

=>

So, the equired tangent is

__Equation of pair of tangents.__

The equation of pair of tangent from (x

_{1},y

_{1}) is

t

^{2}= ss

_{1}

or. (yy

_{1}- 2a(x + x

_{1}))

^{2}= (y

^{2}- 4ax)(y

_{1}

^{2}- 4ax

^{1})

__Why__?

Let p(x

_{1}1 y

_{1}) be point from which tangents is drawin to y

^{2}= 4ax

Let m(h

_{1}R) 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 - x

_{1})

^{2}= (K-y

_{1})(hy

_{1}- K x

_{1})

Putting (x

_{1}y

_{1}) in place of (h

_{1}K) gibes equation of locus of M. as

a(x - x

_{1})

_{2}= (y - y

_{1})(xy

_{1}- yx

_{1})

Which on rearranging gives .

__Illustration 6__.

Find the equation of tangents drawn to y

^{2}+ 12x = 0from point (3,8).

__Ans__Now, y

^{2}= - 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. 3mm

^{2}- 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 +at

^{3}

__Why__?

Slope of normal at (at

^{2}, 2at)

Equation of normal at (at

^{2},2at) is y - 2at = - t (x-at

^{2})

or y = - tx + 2at + at

^{3}

(2) Equation of normal slope m

Putting m = - t in above equation

y = mx - 2am - am

^{3}(3) Equation of normal at (x

_{1},y

_{1})

y-y

_{1}= -y

_{1}/2a (x - x

_{1})

__Some important point__.

(1) Maximum 3 normals can be drawn from anyn piont (h

_{1}K ) to parabola y

^{2}= 4ax .

__Why__?

Equation of normal is

y = - tx + 2at + at

^{3}

So, 3 different value of t are possible when we put (h

_{1}K) in given eqquation

i.r K = - t(n) + 2at + at

^{3}

Hence, 3different normals are possing .

(2) Three distirnt normals can be drawn from point (h

_{1}K ) to parabola y

_{2}4qx iff 4 > 2a .

__Illustration 7__.

Three normal to y

^{2}= 4x pass through point (15,12) Show that one the normal is given by y =x - -3 and find equation of the others.

__Ans__:- y

^{2}= 4x .

=> 4a =4 ie a = 1.

Any normal to parabola is

y = mx - 2am - am

^{3}

putting a= 1 , we get

y = mx - 2am - am

^{3}

As it passses through (15,12) we get

12 = 15m - 2m - m

^{3}

or m

^{3}- 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.