Question

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

• 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

Answer: (D)

$x^2+2xy-3y^2=0$
diff. w.r.t. $x$
$2x+2x\left(y^{\prime }\right)+2y-6y{y}^{\prime }=0$
$2+2y^{\prime }+2-6y^{\prime }=0$
$4y^{\prime }=4$
$y^{\prime }=1$
$\Rightarrow$slope of normal $=-1$
So equation becomes
$y-1=-1(x-1)$
$x+y=2$
solving it with curve
$x^2+2xy-3y^2=0$
$x^2+2x(2-x)-3(2-x{)}^2=0$
$x^2+4x-2x^2-3\left(x^2-4x+4\right)=0$
$-4x^2+16x-12=0$
$x^2-4x+3=0$
$(x-1)(x-3)=0$
$x=1,3$
$\Rightarrow y=1,-1$
thus second point of intersection is $(3,-1)$ is in $4^{\text{th }}$ qud.