JEE 2023 Mathematics PYQ — Let Let be the smallest even value of such that the eccentricity … | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let Hn:1+nx2−3+ny2=1,n∈NLet k be the smallest even value of n such that the eccentricity of Hk is a rational number. If l is the length of the latus rectum of Hk, then find the value of 21l.
Choose the correct answer:
A.
306
(Correct Answer)
B.
360
C.
305
D.
603
Correct Answer:
306
Explanation
Step 1: Eccentricity (e) ki value nikalna
Yahan a2=1+n aur b2=3+n hai.
e=1+1+n3+n
e=1+n1+n+3+n=n+12n+4=n+12(n+2)
Step 2: n ki smallest even value dhundna (taaki e rational ho)
e ko rational hone ke liye, root ke andar wali quantity ek perfect square honi chahiye (q2p2).
Hame n ki even values check karni hain (n=2,4,6,…):
Agar n=2: e=32(4)=38 (Irrational)
Agar n=4: e=52(6)=512 (Irrational)
Agar n=6: e=72(8)=716 (Irrational)
Agar n=14: e=152(16) (Irrational)
Chaliye check karte hain jab 2(n+2)=25 ya numerator/denominator perfect square ban jaye. Ek aisi value n=7 par e=18/8=1.5 hoti, par n even hona chahiye.
Trial and error se: Agar n=49 (odd), nahi chalega.
Sahi value: Agar hum n=31 rakhte to (odd)... Hame even chahiye.
Let's re-examine n+12n+4=q2p2. For n=24 (even): e=2552 (No).
Check n=98: e=99200 (No).
Correction/Shortcut:e2=n+12n+4=2+n+12.
For e to be rational, 2+n+12 must be a square of a rational.
Let 2+n+12=925⟹n+12=97⟹18=7n+7⟹n=11/7 (No).
Let 2+n+12=49⟹n+12=41⟹n+1=8⟹n=7 (Odd).
Let 2+n+12=2549⟹n+12=negative (No).
Let's check n such that n+12(n+2)=1649? No.
Smallest even n that works is n=48:
e=492(50)=49100=710.
10/7 ek rational number hai aur n=48 even hai. To k=48.
Step 3: Latus Rectum (l) nikalna
Hyperbola H48 ke liye:
a2=1+48=49⟹a=7
b2=3+48=51
Length of latus rectum l=a2b2:
l=72×51=7102
Step 4: Final Value
Hame 21l nikalna hai:
21l=21×7102
21l=3×102=306
Explanation
Step 1: Eccentricity (e) ki value nikalna
Yahan a2=1+n aur b2=3+n hai.
e=1+1+n3+n
e=1+n1+n+3+n=n+12n+4=n+12(n+2)
Step 2: n ki smallest even value dhundna (taaki e rational ho)
e ko rational hone ke liye, root ke andar wali quantity ek perfect square honi chahiye (q2p2).
Hame n ki even values check karni hain (n=2,4,6,…):
Agar n=2: e=32(4)=38 (Irrational)
Agar n=4: e=52(6)=512 (Irrational)
Agar n=6: e=72(8)=716 (Irrational)
Agar n=14: e=152(16) (Irrational)
Chaliye check karte hain jab 2(n+2)=25 ya numerator/denominator perfect square ban jaye. Ek aisi value n=7 par e=18/8=1.5 hoti, par n even hona chahiye.
Trial and error se: Agar n=49 (odd), nahi chalega.
Sahi value: Agar hum n=31 rakhte to (odd)... Hame even chahiye.
Let's re-examine n+12n+4=q2p2. For n=24 (even): e=2552 (No).
Check n=98: e=99200 (No).
Correction/Shortcut:e2=n+12n+4=2+n+12.
For e to be rational, 2+n+12 must be a square of a rational.
Let 2+n+12=925⟹n+12=97⟹18=7n+7⟹n=11/7 (No).
Let 2+n+12=49⟹n+12=41⟹n+1=8⟹n=7 (Odd).
Let 2+n+12=2549⟹n+12=negative (No).
Let's check n such that n+12(n+2)=1649? No.
Smallest even n that works is n=48:
e=492(50)=49100=710.
10/7 ek rational number hai aur n=48 even hai. To k=48.
