JEE Main 18 March 2021 Shift 2 Solved Paper

Given,
x2−2y2=4
⇒‌‌‌

x2

4

−‌

y2

2

=1⇒‌

x2

(2)2

−‌

y2

(√2)2

=1
Here, a=2,b=√2
So, Focus (F)=(±ae,0)=(±√6,0)
Now, equation of tangent at P(4,√6) is
4x−2√6y=4‌‌[∵x⋅x−2⋅y⋅y=4,x⋅4−2⋅y⋅√6=4]
⇒‌‌2x−√6y=2...(i)
Putting y=0 in Eq. (i), we get x-intercept of tangent i.e. x=1
∴‌‌Q≡(1,0)
Hence, equation of corresponding latus rectum is x=√6.
∴R≡(√6,‌

2(√6−1)

√6

) [putting x=√6 in Eq. (i), we get y=‌

2(√6−1)

√6

]
∴ Area of △QFR=‌

1

2

×(QF)×(RF)