Question

Find the lengths of tangent, subtangent, normal and subnormal to $y^2=4ax$ at $(a{t}^2,2at)$.

• A. Length of tangent$=2at\sqrt{1-t^2}$; Length of normal$=2a\sqrt{t^2-1}$; Length of subtangent$=2a{t}^2$; Length of subnormal$=2a$
• B. Length of tangent$=2at\sqrt{1+t^2}$; Length of normal$=2a\sqrt{t^2+1}$; Length of subtangent$=2a{t}^2$; Length of subnormal$=2a$
• C. Length of tangent$=2at\sqrt{t^2-1}$; Length of normal$=2a\sqrt{t^2-1}$; Length of subtangent$=2a^2{t}^2$; Length of subnormal$=2a$
• D. None of these

Answer 1

Answer: (B)

Length of tangent$=2at\sqrt{1+t^2}$; Length of normal$=2a\sqrt{t^2+1}$; Length of subtangent$=2a{t}^2$; Length of subnormal$=2a$

Given equation of curve is

$y^2=4ax$

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

Slope of tangent at $(a{t}^2,2at)$ is $={\left(\frac{dy}{dx}\right)}_{(a{t}^2,2at)}=\frac{1}{t}$

Now, the length of tangent at $(a{t}^2,2at)$ is

$=y_1\sqrt{1+{\left(\frac{dx}{dy}\right)}_{(x_1,y_1)}^2}$

$=2at\sqrt{1+t^2}$

$\therefore$ Length of normal at $(a{t}^2,2at)$ is

$=y_1\sqrt{1+{\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}^2}$

$=2at\sqrt{1+\frac{1}{t^2}}$

$=2a\sqrt{t^2+1}$

Length of subtangent = $\frac{y_1}{{\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}}=\frac{2at}{\frac{1}{t}}=2a{t}^2$

Length of subnormal=$y_1{\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=2at.\frac{1}{t}=2a$