Consider a hyperbola $\mathrm{H}: \mathrm{x}^{2}-2 \mathrm{y}^{2}=4$. Let the tangent at a point $\mathrm{P}(4, \sqrt{6})$ meet the x-axis at Q and latus rectum at $\mathrm{R}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{x}_{1}>0$. If F is a focus of H which is nearer to the point P, then the area of $\triangle QFR$ is equal to :

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Hyperbola

Consider a hyperbola

H : x^{2} - 2 y^{2} = 4

. Let the tangent at a point

P (4, 6)

meet the x-axis at Q and latus rectum at

R (x_{1}, y_{1}), x_{1} > 0

. If F is a focus of H which is nearer to the point P, then the area of

△ QF R

is equal to :

Answer: (D)

Writing the hyperbola as

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

gives

a = 2

b^{2} = 2

and

a e = a^{2} + b^{2} = 6

, so the focus nearer to

P (4, 6)

F (6, 0)

. The tangent at

P

\frac{4 x}{4} - \frac{6 y}{2} = 1 ⟹ x - \frac{6}{2} y = 1

Setting

y = 0

gives

Q (1, 0)

; setting

x = 6

, the latus rectum through

F

, gives

y = \frac{2 ( 6 - 1 )}{6} = 2 - \frac{2}{6}

, so

R (6, 2 - \frac{2}{6})

. Since

QF

lies along the

x

-axis and

R

sits vertically above

F

[QF R] = \frac{1}{2} (6 - 1) (2 - \frac{2}{6}) = (6 - 1) (1 - \frac{1}{6}) = 6 + \frac{1}{6} - 2

and

6 + \frac{1}{6} = \frac{7}{6}

, so the area of the triangle is

\frac{7}{6} - 2

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option