JEE Mathematics 2023 Question with Solution

Given: The pair of lines is represented by $2x^2 + xy - 3y^2 = 0$.

Find: The equation of their angle bisectors.

For the homogeneous equation

$$ax^2 + 2hxy + by^2 = 0$$

the combined equation of the angle bisectors is

$$\frac{a-b}{x^2-y^2} = \frac{h}{xy}$$

which is equivalently written as

$$\frac{a-b}{x^2-y^2} = \frac{h}{xy}$$

or

$$\frac{x^2-y^2}{a-b} = \frac{xy}{h}$$

Here,

$$a = 2, \quad h = \frac{1}{2}, \quad b = -3$$

So,

$$\frac{x^2-y^2}{2-(-3)} = \frac{xy}{\frac{1}{2}}$$

that is,

$$\frac{x^2-y^2}{5} = 2xy$$

Hence,

$$x^2-y^2 = 10xy$$

Therefore,

$$x^2-y^2-10xy = 0$$

The correct option from the working is B. Note that this equation matches the listed fourth option text, so the source options/labels are inconsistent.

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

Find: Equation of the angle bisectors.

Compare with

$$ax^2 + 2hxy + by^2 = 0$$

Then,

$$a = 2, \quad 2h = 1 \Rightarrow h = \frac{1}{2}, \quad b = -3$$

Use the standard result:

$$\frac{x^2-y^2}{a-b} = \frac{xy}{h}$$

Substitute the values:

$$\frac{x^2-y^2}{2-(-3)} = \frac{xy}{1/2}$$ $$\frac{x^2-y^2}{5} = 2xy$$ $$x^2-y^2 = 10xy$$ $$x^2-y^2-10xy=0$$

Thus the equation of the angle bisectors is $x^2-y^2-10xy=0$.