Question

# A tangent to the parabola $$y^2=4ax$$ meets the axes at A and B.  Then the locus of mid point of $${AB}$$ is

A
y2+2ax=0
B
y22ax=0
C
y2+ax=0
D
2y2+ax=0

Solution

## The correct option is D $$2y^2+ax=0$$R.E.F.image.A] at $$(x_{1},y_{1})$$Tangent, $$yy_{1} = 2a(x+x_{1})$$$$y(2ax) = 2a (x+ax^{2})$$or $$xy = x+ax^{2}$$x  $$y=\frac{x}{t}+at$$= 0  at$$= -at^{2}$$  0 $$\therefore A = (0,at)$$  $$B(-at^{2},0)$$C is mid point of AB$$\therefore (\frac{0-at^{2}}{2},\frac{0+at}{2}) = (\frac{-at^{2}}{2},\frac{at}{2}) = (h,k)$$at D$$y^{2} = 4ax$$$$= (2at)^{2}= 4a(at^{2})$$$$= (2\times 2k)^{2} = 4a(-2h)$$$$= 16k^{2} = -8ah$$$$= 2k^{2}+ah = 0$$$$\therefore 2y^{2}+ax = 0$$ [locus of midpoint of AB]Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More