JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Hyperbola

Let a line

L_{1}

be tangent to the hyperbola

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

and let

L_{2}

be the line passing through the origin and perpendicular to

L_{1}

. If the locus of the point of intersection of

L_{1}

and

L_{2}

(x^{2} + y^{2})^{2} = α x^{2} + β y^{2}

, then

α + β

is equal to ____.

Answer: 12

Solution

The tangent to the hyperbola

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

at parameter

θ

is:

\frac{x sec θ}{4} - \frac{y tan θ}{2} = 1

Slope of

L_{1}

m_{1} = \frac{s e c θ}{2 t a n θ}

.

Let

(h, k)

be the intersection point. Since

L_{2}

passes through the origin with slope

m_{2} = \frac{k}{h}

and

L_{1} ⊥ L_{2}

m_{1} \cdot m_{2} = - 1 ⟹ \frac{k}{h} \cdot \frac{sec θ}{2 tan θ} = - 1 ⟹ sin θ = - \frac{k}{2 h}

Thus

cos θ = \frac{4 h ^{2} - k ^{2}}{2 h}

.

Since

(h, k)

lies on

L_{1}

\frac{h sec θ}{4} - \frac{k tan θ}{2} = 1

Substituting the expressions for

sec θ

and

tan θ

and simplifying:

h^{2} + k^{2} = 2 4 h^{2} - k^{2}

Squaring:

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

.

Thus

α = 16

β = - 4

, and

α + β = 12

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer