Question

How do you find the equation(s) of the tangent line(s) at the point(s) on the graph of the equation \[{y^2} - xy - 10 = 0\], where \[x = 3\]?

Answer 1

Hint: Here in this question, we have to find the equation of the tangent line, given the x value then find the y value by factoring the given equation, then find the slope of the equation i.e., \[m = \dfrac{{dy}}{{dx}}\] at the point \[\left( {x,y} \right)\], then find the equation of tangent line by using the equation \[y - {y_0} = m\left( {x - {x_0}} \right)\] where \[\left( {{x_0},{y_0}} \right)\] is the point where tangent line passes through.

Complete step by step solution:
The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point.
Consider the given equation when \[x = 3\], then
\[{y^2} - xy - 10 = 0\]------(1)
Substitute \[x = 3\] in the equation (1), then
\[{y^2} - 3y - 10 = 0\]
Now, find the value of y by using the factorization method
Break the middle term as the summation of two numbers such that its product is equal to -10. Calculated above such two numbers are -5 and 2.
\[ \Rightarrow {y^2} + 2y - 5y - 10 = 0\]
Making pairs of terms in the above expression
\[ \Rightarrow \left( {{y^2} + 2y} \right) - \left( {5y + 10} \right) = 0\]
Take out greatest common divisor GCD from the both pairs, then
\[ \Rightarrow y\left( {y + 2} \right) - 5\left( {y + 2} \right) = 0\]
Take \[\left( {y + 2} \right)\] common
\[ \Rightarrow \left( {y + 2} \right)\left( {y - 5} \right) = 0\]
Equate each term to the zero
\[ \Rightarrow \left( {y + 2} \right) = 0\] and \[\left( {y - 5} \right) = 0\]
\[ \Rightarrow y = - 2\] and \[y = 5\]
The points of tangency are \[(3, - 2)\] and \[(3,5)\]
Now, find the slope of the equation at the point of tangency using the first order derivative of given equation i.e., \[m = \dfrac{{dy}}{{dx}}\]
\[ \Rightarrow m = 2y\dfrac{{dy}}{{dx}} - y - x\dfrac{{dy}}{{dx}} = 0\]
On simplification, we get
\[ \Rightarrow \left( {2y - x} \right)\dfrac{{dy}}{{dx}} = y\]
\[ \Rightarrow m = \dfrac{{dy}}{{dx}} = \dfrac{y}{{\left( {2y - x} \right)}}\]
The slope of the equation at the point \[(3, - 2)\] is:
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{(3, - 2)}} = \dfrac{{ - 2}}{{\left( {2\left( { - 2} \right) - 3} \right)}}\]
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{(3, - 2)}} = \dfrac{{ - 2}}{{\left( { - 4 - 3} \right)}}\]
On simplification, we get
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{(3, - 2)}} = \dfrac{{ - 2}}{{ - 7}} = \dfrac{2}{7}\]
And the slope of the equation at the point \[(3,5)\] is:
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{(3,5)}} = \dfrac{5}{{\left( {2\left( 5 \right) - 3} \right)}}\]
On simplification, we get
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{(3,5)}} = \dfrac{5}{{\left( {10 - 3} \right)}} = \dfrac{5}{7}\]
The equation of slope of the tangent line is:
\[y - {y_0} = m\left( {x - {x_0}} \right)\]
the slope of the tangent line at the point \[(3, - 2)\] is:
where \[{x_0} = 3\], \[{y_0} = - 2\] and \[m = \dfrac{2}{7}\] , then
\[ \Rightarrow y - \left( { - 2} \right) = \dfrac{2}{7}\left( {x - 3} \right)\]
\[ \Rightarrow y + 2 = \dfrac{2}{7}x - \dfrac{2}{7}\left( 3 \right)\]
\[ \Rightarrow y + 2 = \dfrac{2}{7}x - \dfrac{6}{7}\]
Subtract both side by 2, then
\[ \Rightarrow y + 2 - 2 = \dfrac{2}{7}x - \dfrac{6}{7} - 2\]
On simplification, we get
\[ \Rightarrow y = \dfrac{2}{7}x - \left( {\dfrac{{6 + 14}}{7}} \right)\]
\[ \Rightarrow y = \dfrac{2}{7}x - \dfrac{{20}}{7}\]
the slope of the tangent line at the point \[(3,5)\] is:
where \[{x_0} = 3\], \[{y_0} = 5\] and \[m = \dfrac{5}{7}\] , then
\[ \Rightarrow y - 5 = \dfrac{5}{7}\left( {x - 3} \right)\]
\[ \Rightarrow y - 5 = \dfrac{5}{7}x - \dfrac{5}{7}\left( 3 \right)\]
\[ \Rightarrow y - 5 = \dfrac{5}{7}x - \dfrac{{15}}{7}\]
Add both side by 5, then
\[ \Rightarrow y - 5 + 5 = \dfrac{5}{7}x - \dfrac{{15}}{7} + 5\]
On simplification, we get
\[ \Rightarrow y = \dfrac{5}{7}x - \left( {\dfrac{{15 - 35}}{7}} \right)\]
\[ \Rightarrow y = \dfrac{5}{7}x - \left( { - \dfrac{{20}}{7}} \right)\]
\[ \Rightarrow y = \dfrac{5}{7}x + \dfrac{{20}}{7}\]
Hence, the equations of the tangent lines at the point of the equation \[{y^2} - xy - 10 = 0\] is \[y = \dfrac{2}{7}x - \dfrac{{20}}{7}\] and \[\,y = \dfrac{5}{7}x + \dfrac{{20}}{7}\].
So, the correct answer is \[y = \dfrac{2}{7}x - \dfrac{{20}}{7}\] and \[\,y = \dfrac{5}{7}x + \dfrac{{20}}{7}\]”.

Note: The concept of the equation of tangent comes under the concept of application of derivatives. Here the differentiation is applied on the equation of line or curve and then substitute the value of x. We should know about the general equation of a line. Hence these types of problems are solved by the above procedure.