JEE Main 2026 — Mathematics Coordinate Geometry

The given equation of the hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$a2x2​−b2y2​=1.

Since it passes through $(6, 4\sqrt{3})$(6,43​), we have:

$$\dfrac{36}{a^2} - \dfrac{48}{b^2} = 1$$a236​−b248​=1

The eccentricity $e$e satisfies $15(e^2 + 1) = 34e$15(e2+1)=34e.

$$15e^2 - 34e + 15 = 0$$15e2−34e+15=0

$$(3e - 5)(5e - 3) = 0$$(3e−5)(5e−3)=0

Since the eccentricity of a hyperbola is $e > 1$e>1, we get $e = \dfrac{5}{3}$e=35​.

We know that $b^2 = a^2(e^2 - 1)$b2=a2(e2−1).

$$b^2 = a^2 \left(\dfrac{25}{9} - 1\right) = \dfrac{16}{9}a^2$$b2=a2(925​−1)=916​a2

Substituting $b^2$b2 into the first equation:

$$\dfrac{36}{a^2} - \dfrac{48}{\dfrac{16}{9}a^2} = 1$$a236​−916​a248​=1

$$\dfrac{36}{a^2} - \dfrac{27}{a^2} = 1$$a236​−a227​=1

$$\dfrac{9}{a^2} = 1 \Rightarrow a^2 = 9$$a29​=1⇒a2=9

Then, $b^2 = \dfrac{16}{9} \times 9 = 16$b2=916​×9=16.

The second hyperbola is given by $\dfrac{x^2}{b^2} - \dfrac{y^2}{2(a^2+1)} = 1$b2x2​−2(a2+1)y2​=1.

Substituting the values of $a^2$a2 and $b^2$b2:

$$\dfrac{x^2}{16} - \dfrac{y^2}{2(9+1)} = 1$$16x2​−2(9+1)y2​=1

$$\dfrac{x^2}{16} - \dfrac{y^2}{20} = 1$$16x2​−20y2​=1

Here, $A^2 = 16 \Rightarrow A = 4$A2=16⇒A=4 and $B^2 = 20$B2=20.

The length of the latus rectum is $\dfrac{2B^2}{A}$A2B2​.

Length of latus rectum $= \dfrac{2 \times 20}{4} = 10$=42×20​=10.

Answer: $10$10