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Hyperbola Triangle Area 4√15: α² = 16 | JEE Main 2026

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

Question
Let H:x2a2−y2b2=1H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 be a hyperbola such that the distance between its foci is 66 and the distance between its directrices is 83\dfrac83. If the line x=αx=\alpha intersects HH at AA and BB such that the area of △AOB\triangle AOB is 4154\sqrt{15}, where OO is the origin, then α2\alpha^2 equals
A1212
B1616correct
C2424
D2525
Solution
Step 1: Distance between foci 2ae=6⇒ae=32ae=6\Rightarrow ae=3. Distance between directrices 2ae=83⇒ae=43\dfrac{2a}{e}=\dfrac83\Rightarrow\dfrac{a}{e}=\dfrac43. Step 2: Multiply: a2=3⋅43=4⇒a=2a^2=3\cdot\dfrac43=4\Rightarrow a=2; divide: e2=34/3=94⇒e=32e^2=\dfrac{3}{4/3}=\dfrac94\Rightarrow e=\dfrac32. Then b2=a2(e2−1)=4(94−1)=5b^2=a^2(e^2-1)=4\left(\dfrac94-1\right)=5. So H:x24−y25=1H:\dfrac{x^2}{4}-\dfrac{y^2}{5}=1. Step 3: At x=αx=\alpha: y=±5(α2−4)4y=\pm\sqrt{\dfrac{5(\alpha^2-4)}{4}}, so A(α,5(α2−4)4)A\left(\alpha,\sqrt{\tfrac{5(\alpha^2-4)}{4}}\right), B(α,−5(α2−4)4)B\left(\alpha,-\sqrt{\tfrac{5(\alpha^2-4)}{4}}\right). Step 4: Area of △AOB=12⋅∣α∣⋅AB=12∣α∣5(α2−4)=415\triangle AOB=\dfrac12\cdot|\alpha|\cdot AB=\dfrac12|\alpha|\sqrt{5(\alpha^2-4)}=4\sqrt{15}. Step 5: Square: 14α2⋅5(α2−4)=16⋅15=240⇒α2(α2−4)=192\dfrac14\alpha^2\cdot5(\alpha^2-4)=16\cdot15=240\Rightarrow\alpha^2(\alpha^2-4)=192. Put u=α2u=\alpha^2: u2−4u−192=0⇒(u−16)(u+12)=0⇒u=16u^2-4u-192=0\Rightarrow(u-16)(u+12)=0\Rightarrow u=16. So α2=16\alpha^2=16. Correct answer: (2)
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