AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

(c) Given curve is x2∕3+y2∕3=a2∕3‌‌......(i)
Now, let a point on curve P(acos3θ,asin3θ)
On differentiating the curve (i) w.r.t x, we get
‌

2

3

x−1∕3+‌

2

3

y−1∕3‌

dy

dx

=0⇒‌

dy

dx

=−(‌

y

x

)1∕3
∴ Slope of tangent at point P is
m=‌

dy

dx

|P=−tan‌θ
∴ Equation of the tangent of the curve at point P is y−asin3θ=−tan‌θ(x−acos3θ)‌‌.....(ii)
∵ The tangent (ii) meets the axes at A and B, so A(a‌cos‌θ,0) and B(0,a‌sin‌θ)
∴ ‌‌AB=√a2cos2θ+a2sin2θ=a