Q108 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Mathematics
If \(T\) is the length of the subtangent drawn at any point on the curve \(3 y^2=4 x^3\) and \(N\) is the length of the subnormal at the same point, then \((\beta T)^2=\)
- A\(4 N^2\)
- B\(4 \mathrm{~N}\)
- Correct answerC\(2 \mathrm{~N}\)
- D\(8 \mathrm{~N}^2\)
Solution
The equation of given curve is
\(3 y^2=4 x^3\) ...(i)
Let a point \(P(h, k)\) on the curve (i), so
\(\left.\frac{d y}{d x}\right|_{(h, k)}=\frac{2 h^2}{k}\)
Now, length of sub-tangent, \(T=\left|k \frac{k}{2 h^2}\right|=\frac{1}{2} \frac{k^2}{h^2}\) and length of sub-normal, \(N=\left|k \frac{2 h^2}{k}\right|=2 h^2\) \(\because \quad(3 T)^2=\frac{9}{4} \frac{k^4}{h^4}=\frac{16 h^6}{4 h^4}=4 h^2=2 N\)
Hence, option (c) is correct.
Hence, the correct answer is option: 'C'.