241)
a+b.dy/dx=0
=>dy/dx=-a/b
now slope of perpendicular line=b/a
x.dy/dx+y=0
=>dy/dx=-y/x=-b/a
=>y/x=b/a
These are few good probs that i've been unable to solve pls solve these for me.
1)If the line ax+by+c=0 is normal to the curve xy+5=0, then a and b have
(a)same sign (b)opposite sign
(c)cannot be discussed (d)none of these
2)What is the equation of the tangent drawn to the curve y2-2x3-4y+8=0 from the point (1,2)?
3)The equation of the tangents to the curve (1+x2)y=1 at the points of its intersection with the curve(x+1)y=1 is what?
4)The equation of the normal at any point θ to the curve x=acosθ+aθsinθ, y=asinθ-aθcosθ is always at a distance of:
(a)2a units from origin (b)a units from origin (c)1/2a units from origin (d)none of these
5)The area bounded by the axes of reference and the normal to y=lnx at(1,0), is:
(a)1 sq unit (b)2 sq unit
(c)1/2 sq unit (d)none of these
Answers urgently reqd
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5 Answers
2424y2-2x3-4y+8=0
=> 2y.dy/dx -6x2-4dy/dx=0
=>dy/dx=6x2/(2y-4)=m
now eqn of tangent:
Y-y=m(X-x)
which passes through (1,2)
243)
easy..forst find pt of intersection and then write eqn of tangent at that point
245)
y=logx
=> dy/dx=1/x
Slope of normal at point (1,0)=-1
eqn:
y=-1(x-1)
=> y+x=1
which forms a triangular area with refrence axis
=> area=(1)(1)/2=1/2
114) Put theta= 0 , we get x=a , y=0
therefore, dy/dx = (dy/ d theta) / (dx / d theta) = tan theta
Eqn of normal :
y - a( sin theta - cos theta) = - cot theta [x -a ( cos theta + theta sin theta)]
We get,
y sin theta - x cos theta = a
Therefore, it is at a constant distance of 'a' from the origin
(b) is correct