Question

The normal to the curve, $x^2+2xy-3y^2=0$, at (1, 1)

• A. does not meet the curve again
• B. meets the curve again in the second quadrant
• C. meets the curve again in the third quadrant
• D. meets the curve again in the fourth quadrant

Answer 1

The correct option is D meets the curve again in the fourth quadrant

$x^2+2xy-3y^2=0$

This is the equation of a pair of straight lines.

$(x-y)(x+3y)=0$

$2x+2y+2x{y}^{\prime }-6y{y}^{\prime }=0$

$\Rightarrow \frac{x+y}{3y-x}=y^{\prime }$

At $(1,1)$

the slope of the tangent is $y^{\prime }=1.$

The slope of the normal is $-1$.

Hence,

The normal has the equation : $x+y=2$

We need to find its intersection with $x+3y=0$

On solving we get,

$2-y+3y=0\Rightarrow 2y=-2$

$y=-1,x=3$

Hence, it meets it in the fourth quadrant.