IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

rini

Prove the locus of the middle points of all tangents drawn from points on the directrix to the parabola y²=4ax is y²(2x + a) = a(3x + a)².

ramyadiamond

Let the tangent be

y= mx +a/m where m is its slope.

Therefore, the point of contact on the parabola will be Q(a/m² , 2a/m)

Take a point P (-a,p) on the directrix from which the tangent is drawn.

Now, it satisfies the eqn. of the tangent,

Hence,

p= -am+a/m

Let the midpoint point of the tangent be M(h,k)

Now,

Using the midpoint formula, we get

p+2a/m = 2k

-am +a/m +2a/m =2k

3a/m -am =2k

am² +2mk-3a=0 -(1)

Also,

a/m²-a =2h

Hence, m² =a/a+2h

Replacing this value of m in eqn. (1), we get

a(a/a+2h) +2k(a/a+2h) -3a=0

2k (a/a+2h) = 3a-a²/a+2h

k(a) = a(a+3h)/(a+2h)

Now, square both the sides,

k²(a) = a²(a+3h)²/(a+2h)

k²= a(a+3h)² / (a+2h)

or, k²(2h+a) = a(a+3h)²

Now replace it by x and y, and u'll get

y²(2x+a) =a(a+3x)²

Hence, proved.

rini

thanks.. ramya

puneet

hii

well done ramya .. take a point from me ...

cheers

ramyadiamond

hey thanks