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

1185116_1277008_ans_0d6e5b5d92aa46bb96e246be8244d258.jpg

Mathematics

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