Question

Find the equation of the tangent line the curve $y={{x}^{2}}-2x+7$ which is,
(a) parallel to the line 2x – y + 9 = 0.
(b) perpendicular to the line 5y – 15x = 13.

Answer 1

Hint: To find the equation of any line, we need a point on this line and the slope of the line. In both the parts, assume a point (x’, y’) on which we will draw the tangent to this curve. Now to find the slope, differentiate the curve with respect to x and put the point (x’, y’) to find the slope at that point. Know, use the information related to the slopes that are given in both the parts of this question and then find the point and the slope and finally, find the equation of the tangent.

Complete step-by-step solution -
Before proceeding with the question, we must know the formula that is required to solve this question.
To find the slope of a tangent at a point (x’, y’) on the curve, we will first differentiate the curve with respect to x to find $\dfrac{dy}{dx}$ and then, we will substitute the point (x’, y’) in $\dfrac{dy}{dx}$.
In the question, we are given a curve $y={{x}^{2}}-2x+7$ and the lines 2x – y + 9 = 0 and 5y – 15x = 13.

Let us assume a point (x’, y’) on this curve. To find the slope of the tangent at this point, we will first find $\dfrac{dy}{dx}$. Differentiating the curve with respect to x, we get,
$\dfrac{dy}{dx}=2x-2$
Substituting (x’, y’) in this equation, we get,
$\dfrac{dy}{dx}=2x'-2$ . . . . . . . . . . . . . (1)
(a) We have to find the equation of the tangent to the curve that is parallel to the line 2x – y + 9 = 0.
For a line ax + by + c = 0, the slope is equal to $\dfrac{-a}{b}$. So, for the line 2x – y + 9 = 0, the slope is equal to $\dfrac{-2}{-1}=2$.
Since this line (of which we have calculated the slope) is parallel to the tangent to the curve, the slope of the tangent to the curve should be equal to 2. From equation (1),
$\begin{aligned} & \Rightarrow 2x'-2=2 \\\ & \Rightarrow x'=2 \\\ \end{aligned}$
Substituting x’ = 2 in the curve, we get y’ equal to,
y’ = ${{2}^{2}}-2\left( 2 \right)+7=4-4+7=7$
So, we have a point on the tangent (2,7) and the slope of this tangent is equal to 2. So, using the slope and point form of the line, the equation of this line is 2x – y + 3 = 0.
(b) We have to find the equation of the tangent to the curve that is perpendicular to the line 5y – 15x = 13 or 15x – 5y + 13 = 0.
For a line ax + by + c = 0, the slope of the perpendicular line is equal to $\dfrac{b}{a}$. So, for the line 15x – 5y + 13 = 0, the slope is equal to $\dfrac{-5}{15}=\dfrac{-1}{3}$.
Since this line (of which we have calculated the slope) is perpendicular to the tangent to the curve, the slope of the tangent to the curve should be equal to $\dfrac{-1}{3}$. From equation (1),
$\begin{aligned} & \Rightarrow 2x'-2=\dfrac{-1}{3} \\\ & \Rightarrow 2x'=\dfrac{5}{3} \\\ & \Rightarrow x'=\dfrac{5}{6} \\\ \end{aligned}$
Substituting x’ = $\dfrac{5}{6}$ in the curve, we get y’ equal to,

& y'={{\left( \dfrac{5}{6} \right)}^{2}}-2\left( \dfrac{5}{6} \right)+7 \\\ & \Rightarrow y'=\left( \dfrac{25}{36} \right)-\left( \dfrac{5}{3} \right)+7 \\\ & \Rightarrow y'=\dfrac{25-60+252}{36} \\\ & \Rightarrow y'=\dfrac{217}{36} \\\ \end{aligned}$$ So, we have a point on the tangent $\left( \dfrac{5}{6},\dfrac{217}{36} \right)$ and the slope of this tangent is equal to $\dfrac{-1}{3}$ . So, using the slope and point form of the line, the equation of this line is $x+3y=\dfrac{227}{12}$. Note: One can also do this question by first finding the parametric form of a point on the parabola given in the question and then finding slope in the form of the parameter used. This method will require a lesser calculation. But one should question this approach only when he/she knows how to find the parametric coordinates of this curve.