TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Given equation of curve is
y2=(2x+1)3. . . (i)
Let P(x1,y1) be any point on Eq. (i)
‌‌ So, ‌‌y12‌‌=(2x1+1)3
⇒‌‌(2x1+1)3∕y12‌‌=1 . . . (ii)
Now, differentiate Eq. (i) w.r.t. x, we get
2y‌

dy

dx

‌‌=3(2x+1)2×2⇒‌

dy

dx

=‌

3(2x+1)2

y

∴(‌

dy

dx

)(x1,y1)‌‌=m=‌

3(2x1+1)2

y1

ST= Length of subtangent
=‌

y1

m

=y1×‌

y1

3(2x1+1)2

=‌

y12

3(2x1+1)2

SN‌‌=‌ length of subnormal ‌=y1m=y1[‌

3(2x1+1)2

y1

]
‌‌=3(2x1+1)2
According to the question,
‌

SN

(ST)2

=‌

3(2x1+1)2

[‌ y12 3(2x1+1)2 ]2

=‌

3(2x1+1)2

1

×‌

9(2x1+1)4

y14

=27[‌

(2x1+1)3

y12

]2=27×1
=27