JEE Main 26-Aug-2021 Shift 1 Solved Paper

Given, equation of ellipse

x2

8

+

y2

4

=1
Then, equation of tangent at (x1,y1) will be

xx1

8

+

yy1

4

=1
Since, tangent is perpendicular to the line x+2y=0, then
(

−x1

2y1

)(

−1

2

)=−1
⇒x1=−4y1
Also,

x12

8

+

y12

4

=1
⇒

16y12

8

+

y12

4

=1 [∵x1=−4y1]
⇒

9

4

y12=1
⇒y12=

4

9

⇒y1=±

2

3

⇒y1=

2

3

[∵(x1,y1). lies in second quadrant]
And x1=−4y1=−4×

2

3

=

−8

3

∴ P(

−8

3

,

2

3

)
Again, a2−b2=a2e2
8−4=8e2
e=

1

√2

Now, S and S' are the foci of the ellipse,
So, S:(ae, 0)
=(2√2.

1

√2

,0)=(2,0)
and S′:(−ae,0)=(−2√2.

1

√2

,0)=(−2,0)
Area of ΔSPS′=

1

2

×4×

2

3

=

4

3

‌[∵ base = 4,height.=

2

3

]
So, (5−e2)‌A=(5−

1

2

)

4

3

=

9

2

.

4

3

=6