Question

The equation of normal at \[\left( at,\dfrac{a}{t} \right)\] to the hyperbola \[xy={{a}^{2}}\] is
(a) \[x{{t}^{3}}-yt+a{{t}^{4}}-a=0\]
(b) \[x{{t}^{3}}-yt-a{{t}^{4}}+a=0\]
(c) \[x{{t}^{3}}+yt+a{{t}^{4}}-a=0\]
(d) \[x{{t}^{3}}+yt-a{{t}^{4}}-a=0\]

Answer 1

Hint:For solving this question you should know about the equation of a normal. When the equation of curve is given by \[y=f\left( x \right)\], then finding the equation of normal line at a curve point \[A\left( {{x}_{1}},{{y}_{1}} \right)\] starts from finding the slope. The slope of the normal to the curve at a point \[A\left( {{x}_{1}},{{y}_{1}} \right)\] is \[A=A=\dfrac{-1}{{{\left( \dfrac{dy}{dx} \right)}_{y={{y}_{1}},x={{x}_{1}}}}}\].

Complete step-by-step solution:
According to the question it is asked to us to finding the equation of normal at a point of the hyperbola and both are given as \[\left( at,\dfrac{a}{t} \right)\] and \[xy={{a}^{2}}\] respectively.
So, for determining the equation of normal:
Step 1: - Determine the equation of the curve \[y=f\left( x \right)\], once that is established, find \[\left( \dfrac{dy}{dx} \right)\] from the given equation of the curve; \[y=f\left( x \right)\].
Step 2: - The next step is very important at it involves calculating the slope for the equation of the normal line to the curve at a point \[A\left( {{x}_{1}},{{y}_{1}} \right)\]. The formula for the slope of the normal is
\[m=\dfrac{-1}{{{\left( \dfrac{dy}{dx} \right)}_{y={{y}_{1}},x={{x}_{1}}}}}\]
Step 3: - The last step is replacing the value of slope in the equation \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]. After solving this you have successfully found the equation for the normal.
Consider, \[xy={{a}^{2}}\]
Differentiating both sides w.r.t x we get,
\[\begin{align}
  & x.\dfrac{dy}{dx}+y=0 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-y}{x} \\
 & \left( {{x}_{1}},{{y}_{1}} \right)=\left( at,\dfrac{a}{t} \right) \\
\end{align}\]
\[\therefore \] Slope of tangent \[m={{\left( \dfrac{dy}{dx} \right)}_{\left( at,\dfrac{a}{t} \right)}}=\dfrac{\dfrac{-a}{t}}{at}=\dfrac{-1}{{{t}^{2}}}\]
\[\therefore \] Equation of normal is
\[\begin{align}
  & y-{{y}_{1}}=\dfrac{-1}{m}\left( x-{{x}_{1}} \right) \\
 & \Rightarrow y-\dfrac{a}{t}=\dfrac{-1}{\left( \dfrac{-1}{{{t}^{2}}} \right)}\left( x-at \right) \\
 & \Rightarrow y-\dfrac{a}{t}={{t}^{2}}\left( x-at \right) \\
 & \Rightarrow yt-a={{t}^{3}}\left( x-at \right) \\
\end{align}\]
Now, if we solve it, then,
\[\Rightarrow x{{t}^{3}}-yt=a{{t}^{4}}-a\]
\[\Rightarrow x{{t}^{3}}-yt-a{{t}^{4}}+a=0\] is equation of normal at \[\left( at,\dfrac{a}{t} \right)\].
\[\therefore \] option ‘(b)’ is the correct answer.

Note: In such a type, we can verify our answers by putting the final answer back to the given situation in our question. Like, here we can put the given coordinates back in the obtained equation, if we get 0 as a result that means our answer is correct and if not then we have to correct our solution.