A normal at `P(x , y)` on a curve meets the x-axis at `Q` and `N` is the foot of the ordinate at `Pdot` If `N Q=(x(1+y^2))/(1+x^2)` , then the equation of curve given that it passes through the point `(3,1)` is (a) `( b ) (c) (d) x^(( e )2( f ))( g )-( h ) y^(( i )2( j ))( k )=8( l )` (m) (b) `( n ) (o) (p) x^(( q )2( r ))( s )+2( t ) y^(( u )2( v ))( w )=11 (x)` (y) (c) `( d ) (e) (f) x^(( g )2( h ))( i )-5( j ) y^(( k )2( l ))( m )=4( n )` (o) (d) None of these

Options

A. `x^(2)-y^(2)=8`
B. `x^(2)+2y^(2)=11`
C. `x^(2)-5y^(2)=4`
D. None of these

Answer

C

Explanation

Equation of normal at point `P(x,y)` is `Y-y=-(dx)/(dy)(X-x)`

`NQ=y(dy)/(dx)=(x(1+y^(2)))/(1+x^(2))`
or `(xdx)/(1+x^(2))=(ydx)/(1+y^(2))`
or `(xdx)/(1+x^(2))=(ydx)/(1+y^(2))`
or `"ln "(1+x^(2))="ln "(1+y^(2))+"ln "c`
or `(1+y^(2))=(1+x^(2))/c`
It passes through (3,1). Thus, `1+1=(1+(3)^(2))/(c)` or c=5
or `(1+y^(2))=(1+x^(2))/(c)`
It passes through (3,1). Thus ,`1+1=(1+(3)^(2))/(c)` or `c=5`
Thus, curve is `5+5y^(2)=1+x^(2)` or `x^(2)-5y^(2)=4`