HyperbolamediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Hyperbola Latus Rectum = 10 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
If the eccentricity ee of the hyperbola x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1, passing through (6,43)\left(6,4\sqrt3\right), satisfies 15(e2+1)=34e15(e^2+1)=34e, then the length of the latus rectum of the hyperbola x2b2−y22(a2+1)=1\dfrac{x^2}{b^2}-\dfrac{y^2}{2(a^2+1)}=1 is
A1010correct
B2020
C2525
D3030
Solution
Step 1: 15(e2+1)=34e⇒15e2−34e+15=015(e^2+1)=34e\Rightarrow15e^2-34e+15=0.
15e2−25e−9e+15=5e(3e−5)−3(3e−5)=(3e−5)(5e−3)=0⇒e=53 or 35.15e^2-25e-9e+15=5e(3e-5)-3(3e-5)=(3e-5)(5e-3)=0\Rightarrow e=\dfrac53\ \text{or}\ \dfrac35.
 e>1⇒e=53e>1\Rightarrow e=\dfrac53. Step 2: e2=1+b2a2⇒b2a2=259−1=169⇒b2=169a2e^2=1+\dfrac{b^2}{a^2}\Rightarrow\dfrac{b^2}{a^2}=\dfrac{25}{9}-1=\dfrac{16}{9}\Rightarrow b^2=\dfrac{16}{9}a^2. Step 3: (6,43)(6,4\sqrt3) on x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1, (43)2=48(4\sqrt3)^2=48:
36a2−48b2=1,48b2=48⋅916a2=27a2.\dfrac{36}{a^2}-\dfrac{48}{b^2}=1,\qquad\dfrac{48}{b^2}=\dfrac{48\cdot9}{16a^2}=\dfrac{27}{a^2}.
⇒36a2−27a2=1⇒9a2=1⇒a2=9, b2=169⋅9=16.\Rightarrow\dfrac{36}{a^2}-\dfrac{27}{a^2}=1\Rightarrow\dfrac{9}{a^2}=1\Rightarrow a^2=9,\ b^2=\dfrac{16}{9}\cdot9=16.
 Step 4: x2b2−y22(a2+1)=1=x216−y220=1\dfrac{x^2}{b^2}-\dfrac{y^2}{2(a^2+1)}=1=\dfrac{x^2}{16}-\dfrac{y^2}{20}=1, since 2(9+1)=202(9+1)=20. Semi-transverse axis =16=4=\sqrt{16}=4.
∴latus rectum=2⋅204=10.\therefore\text{latus rectum}=\dfrac{2\cdot20}{4}=10.
 Correct answer: (1)
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