Question

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

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

Answer 1

The correct option is C meets the curve again in the fourth quadrant.

Given equation of curve is $x^2+2xy-3y^2=0$ .......$\left(1\right)$

On differentiating w.r.t $x$ we get

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

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

At $x=1,y=1,y^{\prime }=1$

${\left(\frac{dy}{dx}\right)}_{(1,1)}=1$

Equation of the normal at $(1,1)$ is

$y-1=-\frac{1}{1}(x-1)$

$\Rightarrow y-1=-x+1$

$\Rightarrow x+y=2$

$\Rightarrow y=2-x$ ......$\left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right)$,we get

$x^2+2x(2-x)-3{(2-x)}^2=0$

$\Rightarrow x^2+4x-2x^2-3(4+x^2-4x)=0$

$\Rightarrow x^2+4x-2x^2-12-3x^2+12x=0$

$\Rightarrow -4x^2+16x-12=0$

$\Rightarrow x^2-4x+3=0$

$\Rightarrow (x-1)(x-3)=0$

$\Rightarrow x=1,x=3$

When $x=1,$then $y=1$ using $\left(2\right)$

When $x=3$ then $y=-1$

$\therefore P(1,1)$ and $Q(3,-1)$

Hence, normal meets the curve again at $(3,-1)$ in fourth quadrant.