Question

Find the equation of tangent and normal to the curve $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ at the point $(\sqrt{2}a,b)$ .

Answer 1

Guess the slope of tangent and normal to the curve and then put it into the general point-slope form of a line i.e. $y-{{y}_{0}}=m(x-{{x}_{0}})$ , where $({{x}_{0}},{{y}_{0}})$ is the point passing through the equation , and hence get the desired equations.

Complete step by step answer:
We know that the tangent line has slope equal to the slope of the given curve
and normal to the curve has slope negative reciprocal of the slope of the given curve.
We will use this information and put the values to the general point –slope form of a line.
Now, we will first evaluate the slope of given curve, i.e. $\dfrac{dy}{dx}$
$\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Differentiating with respect to $''x''$ on both the sides, we get,
[We know the chain rule of differentiation,
$\dfrac{d}{dx}\left( f(g(x)) \right)=f'(g(x)).g'(x)$
So, for differentiating ${{y}^{2}}$ , we will be doing,
$\dfrac{d}{dx}({{y}^{2}})=(2y)\left( \dfrac{dy}{dx} \right)$ ]
$\begin{aligned} & \dfrac{2x}{{{a}^{2}}}-\dfrac{2y}{{{b}^{2}}}\left( \dfrac{dy}{dx} \right)=0 \\\ \end{aligned}$
Now, we can divide by 2 on both the sides of the equation and get,
$\begin{aligned} & \Rightarrow \dfrac{x}{{{a}^{2}}}-\dfrac{y}{{{b}^{2}}}\dfrac{dy}{dx}=0 \\\ & \Rightarrow \dfrac{x}{{{a}^{2}}}=\dfrac{y}{{{b}^{2}}}\dfrac{dy}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{x}{{{a}^{2}}}\times \dfrac{{{b}^{2}}}{y} \end{aligned}$
Slope of the tangent line at $(\sqrt{2}a,b)={{\left. \dfrac{x}{y}\dfrac{{{b}^{2}}}{{{a}^{2}}} \right|}_{(\sqrt{2}a,b)}}$
$=\dfrac{\sqrt{2}a}{b}\dfrac{{{b}^{2}}}{{{a}^{2}}}$
$=\dfrac{\sqrt{2}b}{a}$
Slope of normal at $(\sqrt{2}a,b)=-\dfrac{a}{\sqrt{2}b}$
We know that general point slope form of line is
$y-{{y}_{0}}=m(x-{{x}_{0}})$
where m is the slope of line at $({{x}_{0}},{{y}_{0}})$ and $({{x}_{0}},{{y}_{0}})$ is the point passing through the line.

**So, EQUATION OF TANGENT LINE-
$y-b=\dfrac{\sqrt{2}b}{a}(x-\sqrt{2}a)$
EQUATION OF NORMAL LINE –
$y-b=-\dfrac{a}{\sqrt{2}b}(x-\sqrt{2}a)$ **

Note: The possibility of mistake here is that we could be mistaken with the slope of normal, for that always remember that
Slope of tangent * slope of normal=-1
i.e. Slope of normal to the given curve is the negative reciprocal of the given curve.