JEE Main 2022 — Hyperbola Question with Solution

Equation of L₁ is

$$\frac{x\sec \theta }{4}-\frac{y\tan \theta }{2}=1$$ ..... (i)

Equation of line L₂ is

$$\frac{x\tan \theta }{2}+\frac{y\sec \theta }{4}=0$$ ..... (ii)

$$\because$$ Required point of intersection of L₁ and L₂ is (x₁, y₁) then

$$\frac{x_1\sec \theta }{4}-\frac{y_1\tan \theta }{2}-1=0$$ ...... (iii)

and $$\frac{y_1\sec \theta }{4}-\frac{x_1\tan \theta }{2}=0$$ ...... (iv)

From equations (iii) and (iv)

$$\sec \theta =\frac{4x_1}{x_1^2+y_1^2}$$ and $$\tan \theta =\frac{-2y_1}{x_1^2+y_1^2}$$

$$\therefore$$ Required locus of (x₁, y₁) is

$$(x^2+y^2{)}^2=16x^2-4y^2$$

$$\therefore$$ $$\alpha$$ = 16, $$\beta$$ = $$-$$4

$$\therefore$$ $$\alpha$$ + $$\beta$$ = 12