Question

Find the equation of the tangent and the normal to the following curve at the indicated point.
$x=a\sec t,y=b\tan t$ at $t$.

Answer 1

$x=a\sec t$ and $y=b\tan t$

Differentiating w.r.t $t,$ we get,

$\Rightarrow$ $\frac{dx}{dt}=a\sec t\tan t$ and $\frac{dy}{dt}=b{\sec }^{2}t$

Slope of tangent, $m={\left(\frac{dy}{dx}\right)}_{(t=t)}=\frac{b}{a}cosect$

Now, $(x_1,y_1)=(a\sec t,b\tan t)$

The equation of tangent :

$y-y_1=m(x-x_1)$

$\Rightarrow$ $y-b\tan t=\frac{b}{a}cosect(x-a\sec t)$

$\Rightarrow$ $y-\frac{b\sin t}{\cos t}=\frac{b}{a\sin t}(x-\frac{a}{\cos t})$

$\Rightarrow$ $\frac{y\cos t-b\sin t}{\cos t}=\frac{b}{a\sin t}\left(\frac{x\cos t-a}{\cos t}\right)$

$\Rightarrow$ $y\cos t-b\sin t=\frac{b}{a\sin t}(x\cos t-a)$

$\Rightarrow$ $ay\sin t\cos t-ab{\sin }^{2}t=bx\cos t-ab$

$\Rightarrow$ $bx\cos t-ay\sin t\cos t-ab(1-{\sin }^{2}t)=0$

$\Rightarrow$ $bx\cos t-ay\sin t\cos t=ab{\cos }^{2}t$

Dividing both sides by ${\cos }^{2}t,$

$\Rightarrow$ $bx\sec t-ay\tan t=ab$

Equation of normal is,

$y-y_1=m(x-x_1)$

$\Rightarrow$ $y-b\tan t=\frac{-a}{b}\sin t(x-a\sec t)$

$\Rightarrow$ $\frac{y\cos t-b\sin t}{\cos t}=\frac{-a}{b}\sin t\left(\frac{x\cos t-a}{\cos t}\right)$

$\Rightarrow$ $y\cos t-b\sin t=\frac{-a}{b}\sin t(x\cos t-a)$

$\Rightarrow$ $by\cos t-b^2\sin t=-ax\sin t\cos t+a^2{\sin }^{2}t$

$\Rightarrow$ $ax\sin t\cos t+by\cos t=(a^2+b^2)\sin t$

Dividing both sides by $\sin t,$

$\Rightarrow$ $ax\cos t+by\cot t=a^2+b^2$

$\therefore$ The equation of tangent is $bx\sec t-ay\tan t=ab$.

$\therefore$ The equation of normal is $ax\cos t+by\cot t=a^2+b^2$