PQ is chord of hyperbola x^24 - y^2b^2 = 1 perpendicular to x-axis. OPQ is equilateral (e= 3). Area OPQ is

Given: Hyperbola x^24 - y^2b^2 = 1 and eccentricity e= 3. The chord PQ is perpendicular to the x-axis, so it is a vertical chord. Also, OPQ is equilateral. Find: Area of OPQ. Using the relation for hyperbola, b^2 = a^2(e^2-1) Here, a^2=4 and e^2=3. Therefore, b^2 = 4(3-1)=8 So the hyperbola becomes x^24 - y^28 = 1 Since PQ is a vertical chord, let its equation be x=k. If point P is (k,y1), then by symmetry point Q is (k,-y1). Because OPQ is equilateral with vertex at the origin, the line OP makes an angle of 30^ with the x-axis. Hence, y1k= 30^ = 1 3 So, k^2 = 3y1^2 Substitute x=k and y=y1 into the hyperbola: k^24 - y1^28 = 1 Using k^2=3y1^2, 3y1^24 - y1^28 = 1 6y1^2-y1^28=1 5y1^28=1 y1^2= 85 The side of the equilateral triangle is PQ = 2y1 Area of an equilateral triangle is 34( side)^2 Thus, Area = 34(2y1)^2 = 3y1^2 Area = 3 85 = 8 35 Therefore, the correct option is C.

PQ is chord of hyperbola x^24 - y^2b^2 = 1 perpendicular to x-axi… — Solution | JEE Mathematics

MCQMediumJEE 2026Conic Sections (Parabola, Ellipse, Hyperbola)

PQ is chord of hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$ perpendicular to x-axis. $\triangle OPQ$ is equilateral ( $e=\sqrt{3}$ ). Area OPQ is

A
$\frac{11}{5}$
B
$\frac{9}{5}$
C
$\frac{8\sqrt{3}}{5}$
D
$2\sqrt{3}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$ and eccentricity $e=\sqrt{3}$ . The chord $PQ$ is perpendicular to the $x$ -axis, so it is a vertical chord. Also, $\triangle OPQ$ is equilateral.

Find: Area of $\triangle OPQ$ .

Using the relation for hyperbola,

b^2 = a^2(e^2-1)

Here, $a^2=4$ and $e^2=3$ . Therefore,

b^2 = 4(3-1)=8

So the hyperbola becomes

\frac{x^2}{4} - \frac{y^2}{8} = 1

Since $PQ$ is a vertical chord, let its equation be $x=k$ . If point $P$ is $(k,y_1)$ , then by symmetry point $Q$ is $(k,-y_1)$ .

Because $\triangle OPQ$ is equilateral with vertex at the origin, the line $OP$ makes an angle of $30^\circ$ with the $x$ -axis. Hence,

\frac{y_1}{k}=\tan 30^\circ=\frac{1}{\sqrt{3}}

So,

k^2 = 3y_1^2

Substitute $x=k$ and $y=y_1$ into the hyperbola:

\frac{k^2}{4} - \frac{y_1^2}{8} = 1

Using $k^2=3y_1^2$ ,

\frac{3y_1^2}{4} - \frac{y_1^2}{8} = 1

\frac{6y_1^2-y_1^2}{8}=1

\frac{5y_1^2}{8}=1

y_1^2=\frac{8}{5}

The side of the equilateral triangle is

PQ = 2y_1

Area of an equilateral triangle is

\frac{\sqrt{3}}{4}(\text{side})^2

Thus,

\text{Area} = \frac{\sqrt{3}}{4}(2y_1)^2 = \sqrt{3}y_1^2

\text{Area} = \sqrt{3}\cdot \frac{8}{5} = \frac{8\sqrt{3}}{5}

Therefore, the correct option is C.

Common mistakes

Assuming the chord $PQ$ is horizontal because it is perpendicular to the $x$ -axis. This is wrong because a line perpendicular to the $x$ -axis is vertical. Take the chord as $x=k$ , not $y=k$ .
Using the ellipse relation for eccentricity instead of the hyperbola relation. For a hyperbola, $b^2=a^2(e^2-1)$ . Using any other relation gives the wrong value of $b^2$ .
Taking $\angle POX=60^\circ$ instead of $30^\circ$ . In an equilateral triangle with base $PQ$ vertical and origin as the third vertex, the median from $O$ to $PQ$ lies along the $x$ -axis, so each side makes $30^\circ$ with the $x$ -axis.

Practice more Conic Sections (Parabola, Ellipse, Hyperbola) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Conic Sections (Parabola, Ellipse, Hyperbola)

Answer

Step-by-step solution

Common mistakes

Practice more Conic Sections (Parabola, Ellipse, Hyperbola) questions

Related questions