Question

Find the equation of the tangent and normal to the curve \[16{x^2} + {y^2} = 145\] at the point \[({x_1},{y_1})\] , where \[{x_1} = 2\] and \[{y_1} > 0\].

Answer 1

Hint: To solve this question, first find the point of intersection of tangent and normal to the curve at the point. Once we find the point of intersection, we can find the slope of tangent and normal at that point. We can use the equations of straight lines to find the equation of tangent and normal at that point.

Complete step-by-step solution:
 It is given that an equation of ellipse \[16{x^2} + {y^2} = 145\] and a point on it \[({x_1},{y_1})\] where \[{x_1} = 2\] and \[{y_1} > 0\]
Let’s start with Finding the equation of the curve.
From the question, we can see that the equation of the curve is given to be \[16{x^2} + {y^2} = 145\] . So, this is the equation we will work on.
Now Find the value of \[{x_1}\] and \[{y_1}\]
It is already given that \[{x_1} = 2\] and \[{y_1} > 0\] .
We can find the value of \[{y_1}\] by substituting the value of \[{x_1}\] in the given equation.
Substituting the value of \[{x_1}\] in the given equation, we have,
\[\begin{array}{l}16{x^2} + {y^2} = 145\\ \Rightarrow 16 \times {2^2} + {y^2} = 145\\ \Rightarrow 64 + {y^2} = 145\\ \Rightarrow {y^2} = 81\\ \Rightarrow y = \pm 9\end{array}\]
Since, it is given that \[{y_1} > 0\] , \[y = 9\]
Therefore, \[({x_1},{y_1}) = (2,9)\]
Finding the slope of tangent at \[({x_1},{y_1}) = (2,9)\]
To find the equation of tangent at \[({x_1},{y_1}) = (2,9)\] , we first have to find the slope of the tangent at that point. Slope can be founded by finding the derivative of the equation at that point.
Let us find the derivative of \[16{x^2} + {y^2} = 145\] at \[({x_1},{y_1}) = (2,9)\] ,
Differentiating the above function with respect to x, we get,
\[\begin{array}{l} \Rightarrow \dfrac{{dy}}{{dx}}(16{x^2} + {y^2} = 145)\\ \Rightarrow \dfrac{{dy}}{{dx}}(16{x^2} + {y^2}) = \dfrac{{dy}}{{dx}}(145)\\ \Rightarrow 32x + 2y\dfrac{{dy}}{{dx}} = 0\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 32x}}{{2y}}\end{array}\]
Now, we will substitute \[({x_1},{y_1}) = (2,9)\] in the equation. We get,
\[\begin{array}{l} \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 32x}}{{2y}}\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 32 \times 2}}{{2 \times 9}}\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 32}}{9}\end{array}\]
This is the slope of the tangent and can be written as \[{m_t} = \dfrac{{ - 32}}{9}\]
Finding the equation of tangent,
Equation of a tangent with slope \[{m_t}\] at point \[({x_1},{y_1})\] is given as
\[y - {y_1} = {m_t}(x - {x_1})\]
Substituting the value of \[{m_t}\]and \[({x_1},{y_1})\] in the above equation, we get,
\[\begin{array}{l}y - 9 = \dfrac{{ - 32}}{9}(x - 2)\\ \Rightarrow y - 9 = \dfrac{{ - 32x + 64}}{9}\\ \Rightarrow 9y - 81 = - 32x + 64\\ \Rightarrow 9y + 32x = 145\end{array}\]
Hence, the equation of tangent at \[({x_1},{y_1}) = (2,9)\] is \[9y + 32x = 145\] .
Finding the slope of normal,
We know that normal to the tangent, is perpendicular to it. Slope of normal is given by:
\[{m_n} \times {m_t} = - 1\]
Substituting the value of \[{m_t} = \dfrac{{ - 32}}{9}\] , we get
\[\begin{array}{l}{m_n} \times {m_t} = - 1\\ \Rightarrow {m_n} \times \dfrac{{ - 32}}{9} = - 1\\ \Rightarrow {m_n} = \dfrac{9}{{32}}\end{array}\]
Finding the equation of normal,
Equation of a normal with slope \[{m_n}\] at point \[({x_1},{y_1})\] is given as
\[y - {y_1} = {m_n}(x - {x_1})\]
Substituting the value of \[{m_n}\] and \[({x_1},{y_1})\] in the above equation, we get,
\[\begin{array}{l}y - {y_1} = {m_n}(x - {x_1})\\ \Rightarrow y - 9 = \dfrac{9}{{32}}(x - 2)\\ \Rightarrow y - 9 = \dfrac{{9x - 18}}{{32}}\\ \Rightarrow 32y - 288 = 9x - 18\\ \Rightarrow 32y - 9x = 270\end{array}\]

Hence, the equation of normal at \[({x_1},{y_1}) = (2,9)\] Is \[32y - 9x = 270\].

Note: To solve these types of questions, you must remember the equations of straight lines and try to find the parameters used for calculating the equations. Here, we had to find the point of intersection and gradient at that point and if the line is perpendicular the slope is equal to \[ - 1\] .