Question
The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0,$$ at $$\left( {1,\,1} \right)$$
A.
meets the curve again in the third quadrant.
B.
meets the curve again in the fourth quadrant.
C.
does not meet the curve again.
D.
meets the curve again in the second quadrant.
Answer :
meets the curve again in the fourth quadrant.
Solution :
Given curve is
$${x^2} + 2xy - 3{y^2} = 0.....(1)$$
Differentiate w.r.t. $$x,\,\,\,2x + 2x\frac{{dy}}{{dx}} + 2y - 6y\frac{{dy}}{{dx}} = 0$$
$${\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,\,1} \right)}} = 1$$
Equation of normal at $$\left( {1,\,1} \right)$$ is
$$y = 2 - x.....(2)$$
Solving equation (1) and (2), we get $$x=1 ,\,3$$
Point of intersection $$\left( {1,\,1} \right),\,\left( {3,\, - 1} \right)$$
Normal cuts the curve again in 4th quadrant.