Question

Find the lengths of tangent, subtangent, normal and

subnormal to y² = 4ax at (at², 2at) –

• A. 2at$\sqrt{1 + t^{2}}$, 2a$\sqrt{t^{2} + 1}$, 2at², 2a
• B. 2a$\sqrt{t^{2} + 1}$, 2at², 2a, 2at$\sqrt{1 + t^{2}}$
• C. 2$\sqrt{t^{2} + 1}$, 2at$\sqrt{1 + t^{2}}$, 2a, 2a t²
• D. None of these

Answer 1

Answer: (A)

2at$\sqrt{1 + t^{2}}$, 2a$\sqrt{t^{2} + 1}$, 2at², 2a

Answer 2

We have the given curve,

y² = 4ax … (i)

Differentiating equation (i) both sides w.r.t. x, we get

2y $\frac{dy}{dx}$ = 4a

$\left\lbrack \frac{dy}{dx} \right\rbrack_{(at^{2},2at)}$ = $\frac{4a}{4at}$ = $\frac{1}{t}$ … (ii)

Now, the length of tangent at (at², 2at) is

= $y_{1}\sqrt{1 + \left( \frac{dx}{dy} \right)_{(x_{1},y_{1})}^{2}}$

Ž = 2at $\sqrt{1 + t^{2}}$

[using (ii)]

\ Length of normal at (at², 2at) is

Ž = $y_{1}\sqrt{1 + \left( \frac{dy}{dx} \right)_{(x_{1},y_{1})}^{2}}$

Ž = 2at $\sqrt{1 + 1/t^{2}}$

Ž = 2a $\sqrt{t^{2} + 1}$

\ Length of subtangent Ž $\frac{y_{1}}{\left\lbrack \frac{dy}{dx} \right\rbrack_{(x_{1},y_{1})}}$= $\frac{2at}{1/t}$ = 2at²

Length of subnormal Ž y₁$\left\lbrack \frac{dy}{dx} \right\rbrack_{(x_{1},y_{1})}$ = 2at . $\frac{1}{t}$ = 2a.