AP EAPCET 05-Jul-2022 Shift 1 Paper

Given curve 3y=6x−5x3
⇒‌‌‌

dy

dx

=2÷5x2
Slope of normal =‌

−1

2−5x2

Equation of normal with slope ‌

−1

2−5x2

passing through origin (0,0).
‌

(y−y1)

x−x1

=m
⇒‌

y−0

x−0

=‌

−1

2−5x2

⇒‌

y

x

=‌

−1

2−5x2

Since, (h,k) lies on normal
‌

k

h

‌=‌

−1

2−5h2

⇒‌

k

h

‌=‌

1

5h2−2

.....(i)
Since, (h,k) lies on curve
‌3k=6h−5h3‌. ‌
‌⇒‌‌3k=h(6−5h2)
‌⇒‌‌‌

k

h

=‌

6−5h2

3

‌ .....(ii)
On solving Eqs. (i) and (ii),
‌‌

1

5h2−2

=‌

6−5h2

3

⇒‌‌3=(6−5h2)(5h2−2)
⇒‌‌3=30h2−12−25h4+10h2
⇒15=−25h4+40h2
⇒5h4−8h2+3=0
For simplicity, let h2=z
The equation becomes 5z2−8z+3=0
(5z−3)(z−1)=0
Now, z=‌

3

5

,1
Putting h2=z
⇒‌‌h2=‌

3

5

⇒h=±√‌

3

5

It is not in option.
or (z−1)=0
z=1
Putting ⋅h2=z
⇒‌‌h2=1
h=±1
But h=−1 is not in option.
∴h=1 is correct