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Question

If lines xy+2=0 and 2xy2=0 meet at a point P, then equation of tangent drawn to the parabola y2=8x from the point P is

A
x2y+8=0
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B
x+y16=0
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C
3xy16=0
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D
x3y+16=0
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Solution

The correct option is A x2y+8=0
xy+2=02xy2=0}
Solving the two equations, we get
x=4 and y=6
So, the tangent passes through (4,6)
Let slope of the tangent be m.
y6=m(x4)
y=mx+64m which is tangent to the parabola y2=8x
64m=2m [Using c=am]
3m2m2=1
2m23m+1=0
m=12,1

For m=12,
y=12x+62
x2y+8=0

For m=1,
y=x+2 that is the given line itself.
So, tangent is :x2y+8=0

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