Hyperbola JEE 2024: Latus Rectum, Scale and Answer 30

What does the hyperbola JEE 2024 latus-rectum question actually determine?

A horizontal hyperbola has a latus rectum through its right focus, and the segments joining its endpoints to the centre meet at a right angle. The hyperbola JEE 2024 question has official answer 30 under an implicit normalisation: the angle fixes the ratio of the squared semi-axes, not their absolute size.

The hyperbola opening left and right with coordinate axes x and y, centre O(0,0), right vertex A(a,0), right focus F(ae,0), vertical latus rectum PQ with P(ae,b²/a) above and Q(ae,−b²/a) below F, and segments OP and OQ enclosing a marked right angle at O.

The given hyperbola is

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .

Identify the coprime positive integers and the radicand in the supplied form, then calculate their squared sum:

b^{2} = \frac{l}{m} (1 + \sqrt{n}), l^{2} + m^{2} + n^{2} = ?

This is a numerical-answer question, with no answer options supplied. Its Hard question-bank tag does not measure observed student performance.

Source note: 2024 is the supplied question-bank label; its date field reads 2023-01-24, or 24 January 2023. These fields do not establish confirmed 2024 shift provenance.

How does the right angle become a dot-product equation?

The position vectors of the endpoints are perpendicular, so their dot product is zero. First derive the endpoints: the focus is the midpoint of the latus rectum, not one of its ends.

For this horizontal hyperbola,

a > 0, b > 0, e = \sqrt{1 + \frac{b^{2}}{a^{2}}} > 1, b^{2} = a^{2} (e^{2} - 1) .

The right focus and the line containing its latus rectum are

F (a e, 0), x = a e .

Substitute this horizontal coordinate into the hyperbola:

e^{2} - \frac{y^{2}}{b^{2}} = 1,

y^{2} = b^{2} (e^{2} - 1) = \frac{b^{4}}{a^{2}}, y = \pm \frac{b^{2}}{a} .

Hence the endpoints are

P (a e, \frac{b^{2}}{a}), Q (a e, - \frac{b^{2}}{a}) .

Each vertical coordinate has magnitude equal to half the full latus-rectum length. The full length is

P Q = \frac{2 b^{2}}{a} .

Apply the right-angle condition to the endpoint vectors, not to vectors drawn from the focus:

\vec{O P} \cdot \vec{O Q} = 0,

(a e) (a e) + (\frac{b^{2}}{a}) (- \frac{b^{2}}{a}) = 0 .

The opposite signs of the vertical coordinates give

a^{2} e^{2} - \frac{b^{4}}{a^{2}} = 0 ⟹ a^{4} e^{2} = b^{4} .

How do you solve for eccentricity and reject the invalid root?

Only the larger quadratic root is valid because a hyperbola’s eccentricity exceeds one. A real root alone is not enough.

Square the earlier relation between the semi-axes and eccentricity:

b^{4} = a^{4} (e^{2} - 1)^{2} .

Substitute into the dot-product result and divide by the strictly positive fourth power:

a^{4} e^{2} = a^{4} (e^{2} - 1)^{2}, a^{4} > 0,

$e^{2} = (e^{2} - 1)^{2} .$ Set $u = e^{2} > 1 .$

Expanding and collecting terms gives

u = u^{2} - 2 u + 1, u^{2} - 3 u + 1 = 0 .

Apply the quadratic formula:

u = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} .

Reject the smaller root because it cannot represent the squared eccentricity of a hyperbola:

\frac{3 - \sqrt{5}}{2} < 1 .

Retaining the larger root gives

e^{2} = \frac{3 + \sqrt{5}}{2},

\frac{b^{2}}{a^{2}} = e^{2} - 1 = \frac{1 + \sqrt{5}}{2} .

This ratio is the result determined by the geometry. No absolute semi-axis length has entered the calculation.

Why is the official answer 30, and where did the squared semi-axis go?

The official coefficient matching treats the radical expression as the squared-axis ratio, equivalently normalising the squared transverse semi-axis to one. Neither condition is explicit in the supplied question text.

The actual consequence of the geometry is

b^{2} = \frac{a^{2}}{2} (1 + \sqrt{5}) .

Enlarging both semi-axes by the same positive factor preserves the angle and their squared ratio. It changes the absolute squared conjugate semi-axis:

a^{'} = k a, b^{'} = k b, k > 0,

\frac{b^{' 2}}{a^{' 2}} = \frac{b^{2}}{a^{2}}, b^{' 2} = k^{2} b^{2} .

Thus the literal statement leaves the scale unspecified. The official matching effectively uses

\frac{b^{2}}{a^{2}} = \frac{l}{m} (1 + \sqrt{n}),

or, equivalently for its displayed expression, the normalisation $a^{2} = 1 .$

Under that stated interpretation,

l = 1, m = 2, n = 5, gcd (1, 2) = 1 .

Therefore,

l^{2} + m^{2} + n^{2} = 1^{2} + 2^{2} + 5^{2} = 1 + 4 + 25 = 30 .

30 is the official answer under this interpretation. The right angle does not prove the scale assumption.

How can a sign error in the square produce a wrong numerical answer?

Changing the squared constant from positive to negative produces a different quadratic that still has an admissible-looking root. This is a constructed wrong numerical-answer route, not an actual option or a documented student response.

Compare the correct expansion with the incorrect one:

(u - 1)^{2} = u^{2} - 2 u + 1 (correct),

(u - 1)^{2} \neq u^{2} - 2 u - 1 (incorrect) .

The incorrect route proceeds as follows:

u = u^{2} - 2 u - 1 ⟹ u^{2} - 3 u - 1 = 0 ⟹ u = \frac{3 + \sqrt{13}}{2} .

The same coefficient matching would then wrongly give

l = 1, m = 2, n = 13,

$1^{2} + 2^{2} + 13^{2} = 174 .$ The squared constant must be positive. Checking only that eccentricity exceeds one cannot detect a wrongly derived quadratic.

Check the expansion at one: the correct square gives zero, but the incorrect expression does not.

u = 1 : (u - 1)^{2} = 0, u^{2} - 2 u - 1 = - 2 .

How do you apply these checks to two related coordinate-geometry questions?

A specified semi-axis fixes the scale; the sign of the endpoint dot product determines the angle type. Both questions below are original practice from the same chapter, not additional verified PYQs.

Question 1: For the following hyperbola, a latus rectum subtends a right angle at the centre. Find the squared conjugate semi-axis.

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

Worked answer: Use the derived ratio with the specified scale:

a^{2} = 4, \frac{b^{2}}{a^{2}} = \frac{1 + \sqrt{5}}{2} .

b^{2} = 4 (\frac{1 + \sqrt{5}}{2}) = 2 (1 + \sqrt{5}) .

Unlike the supplied question, this equation fixes the squared transverse semi-axis. The absolute value is therefore determined.

Question 2: Find the endpoints of the right latus rectum and decide whether their angle at the centre is acute, right or obtuse:

\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1 .

Worked answer: Calculate the focus coordinate before writing the endpoints:

a = 3, b^{2} = 16, e = \sqrt{1 + \frac{16}{9}} = \frac{5}{3}, a e = 5 .

P (5, \frac{16}{3}), Q (5, - \frac{16}{3}) .

Their endpoint-vector dot product is

\vec{O P} \cdot \vec{O Q} = 25 - \frac{256}{9} = - \frac{31}{9} < 0 .

The angle is obtuse. Before submitting a related answer, make two separate checks: confirm that the scale is specified for an absolute squared semi-axis, then check the endpoint dot-product sign for the angle type.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

For a worked example of the same idea, see Thermodynamics JEE 2017: Graphite–Diamond Equilibrium.

Frequently asked questions

How do I find the latus rectum endpoints of a hyperbola?

For x²/a² − y²/b² = 1, the right focus is (ae, 0), so its latus rectum lies on x = ae. Substituting this into the hyperbola gives the endpoints (ae, b²/a) and (ae, −b²/a). The full latus-rectum length is 2b²/a.

What happens when a hyperbola's latus rectum subtends a right angle at the centre?

The position vectors of its endpoints have zero dot product, giving a⁴e² = b⁴. Combining this with b² = a²(e² − 1) gives e² = (3 + √5)/2 and b²/a² = (1 + √5)/2. The angle fixes this ratio, not the absolute semi-axis lengths.

Why is the official answer to the hyperbola JEE 2024 question 30?

The official coefficient matching implicitly uses b²/a² = (l/m)(1 + √n), or equivalently normalises a² to 1 in the displayed expression for b². This gives l = 1, m = 2 and n = 5, so l² + m² + n² = 30. The supplied wording does not explicitly fix that scale.

Why do we reject the smaller root when finding eccentricity?

A hyperbola requires e > 1, so u = e² must also exceed 1. The root (3 − √5)/2 is less than 1 and must be rejected. The admissible root is (3 + √5)/2.