JEE Main 25 Feb 2021 Shift 2 Solved Paper

If the curves cut at right angle, then product of slopes will be −1.
First curve x=y4
Differentiate it, we get
1‌‌=4y3‌

dy

dx

‌

dy

dx

‌‌=‌

1

4y3

⇒‌‌‌ Slope of first curve ‌(m1)‌‌=‌

1

4y13

[at point (x1,y1) ]
Second curve xy=k
Differentiate it, 0=x‌

dy

dx

+y
⇒‌‌‌

dy

dx

=‌

−y

x

Slope of second curve (m2)=‌

−y1

x1

[at (x1,y1) ]
⇒‌‌m1⋅m2=−1
⇒‌‌‌

1

4y13

(‌

−y1

x1

)=−1⇒‌

−1

4y12x1

=−1
⇒‌‌‌

−1

4(y1)6

=−1
⇒‌‌y16=‌

1

4

Also, x1y1=k, using x1=y14, we get k=y15 or k6=(y1)30
∵‌‌y16=‌

1

4

, then y130=(‌

1

4

)5
∴(4k)6=46⋅k6=46(y1)30=46(‌

1

4

)5=4
∴(4k)6=4