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Parabola : part 1 ques?
Q1.If a line x +y =1 cut the parabola y^2 = 4ax in the points A and B and normals drawn at A and B meet at C. The normal to the parabola from C ,other than above two meet the parabola in D, then find D
1 Answer
- ?Lv 71 decade agoFavorite Answer
The intersection of the line and parabola is the points
( 2a+1 -+ 2sqrt(a^2+a) , -2a +- 2sqrt(a^2+a) )
The derivative 2y dy/dx = 4a implies the normal slope is -2y / 4a = -y/ 2a.
The equations of the normals are
y = -[-2a + 2sqrt(a^2+a)]/2a * [x - ( 2a+1 - 2sqrt(a^2+a) ) ] -2a + 2sqrt(a^2+a)
and
y = -[-2a - 2sqrt(a^2+a)]/2a * [x - ( 2a+1 + 2sqrt(a^2+a) ) ] -2a - 2sqrt(a^2+a)
These intersect when x=6a+1 and y = -2.
So C(6a+1,-2) must be on another normal to y^2=4ax at D(y^2 / 4a, y).
Then the slope of the normal at D must equal the slope of CD:
-y/2a = (y+2) / ( y^2 /(4a) -6a -1).
Then y=4a and x=4a. So D is (4a,4a).
Source(s): I used wolframalpha.com for some of the algebra.
