Q32 / 90 · JEE Main 2015 (04 Apr) · Mathematics

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

Solution

Given

x^{2} + 2 x y - 3 y^{2} = 0

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

Pair of straight lines passing through the origin.

∵ x + 3 y = 0

x - y = 0

Normal exists at

(1, 1)

which is on

x - y = 0

\Rightarrow

The slope of normal at

(1, 1) = - 1

∴

Equation of normal will be

y - 1 = - (x - 1)

[y - y_{1} = m (x - x_{1})]

y - 1 = - x + 1

x + y = 2

Now, find the point of intersection with

x + 3 y = 0

\begin{matrix} x + y = 2 \\ x + 3 y = 0 \end{matrix}

----------------

- 2 y = 2

\Rightarrow y =-1, x =3

∵ (3, - 1)

lies in the fourth quadrant.

Hence, the correct answer is option: 'A'.