IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

kane

find the equation of common tangents to x²+y²-6y+4=0 and y²=x?

ans:x-2y+1=0

plz solve all the steps,no need to ask for rates.rates are always guaranteed from my side.

kane

help yaar

rhd92781

eqn of tangent to parabola y^2=4ax is y=mx+a/m
here a=1/4
so tang is y=mx+1/4m (m is slope) or 4m^2x-4my+1=0
it is also tangent to circle whose centre is (0, 3) radius = root(5)
so dist of tang frm (0, 3) = root(5)
so (1-12m)/ (4mroot(m^2+1)) = root(5)
solving it u'll get values of m and common tangent

kane

thnxs buddy,kindly see my article on thermodynamics

Sivvar

the equations given are those of a circle n a parabola as shown in the dia:

first let tangent be y= mx+c

c=1/4m if this line is tangent to the parabola..

next frm the eqn of circle we find the centre to be (0,3)..radius is root 5.

dist of centre frm tangent=radius.

usin this n eqn of tangent

(3 - 1/4 m) / ((root) (1+m²)) = (root) 5

square this..

80m⁴- 64 m²+24m -1=0;

big eqn:)....by inspection u find this root is 1/2..

then eqn of tangent is y -1/2 x - 1/2=0.

n thts the ans u gave kane..hope its clear..

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