Answer
384.9k+ views
Hint: Here we will first assume the coordinates of the point at which the tangent is drawn to be any variable. We will find the slope of the tangent using the assumed point and then we will find the slope of the line joining the two given points on the curve. Then we will equate both the slope and simplify further to get the value of the abscissa of the point of the tangent.
Formula used:
If \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] are two points, then the slope of the line joining the two points is given by \[{\rm{Slope}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].
Complete step by step solution:
Here we need to find the abscissa of the point on the curve at which the tangent is drawn.
The given curve is \[y = 3{x^2} - 11x - 15\].
Let the coordinates of the point at which the tangent is drawn be \[\left( {a,b} \right)\].
Now, we will draw a rough diagram to show the given information.
Now, we will find the slope of the tangent on the given curve. For that, we will first find the derivative of the given equation of the curve.
We will differentiate the equation of the given curve with respect to \[x\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {3{x^2} - 11x - 15} \right)\]
On differentiating each term, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 6x - 11\]
Now, to find the slope of the tangent, we will substitute the value of the coordinates of the point of the tangent in the above equation.
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {a,b} \right)}} = 6a - 11\]
As it is given that the line joining the two given points i.e. \[\left( {5,5} \right)\] and \[\left( {11,227} \right)\] which lies on the curve and the tangent are parallel to each other, then the slope of the line and that tangent will be equal.
Now, we will calculate the slope of the line joining the points \[\left( {5,5} \right)\] and \[\left( {11,227} \right)\].
Substituting \[{x_1} = 5,{y_1} = 5,{x_2} = 11\] and \[{y_2} = 227\] in the formula \[{\rm{Slope}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\], we get
\[{\rm{Slope}} = \dfrac{{227 - 5}}{{11 - 5}}\]
Subtracting the terms, we get
\[ \Rightarrow {\rm{Slope}} = \dfrac{{222}}{6}\]
Dividing 222 by 6, we get
\[ \Rightarrow {\rm{Slope}} = 37\]
Now, we will equate the slope of the line and the slope of the tangent. Therefore, we get
\[6a - 11 = 37\]
Adding 11 on both sides, we get
\[ \Rightarrow 6a - 11 + 11 = 37 + 11\]
\[ \Rightarrow 6a = 48\]
Dividing both sides by 6, we get
\[ \Rightarrow \dfrac{{6a}}{6} = \dfrac{{48}}{6}\]
\[ \Rightarrow a = 8\]
Hence, the required abscissa of the point is equal to 8.
Therefore, the correct answer is option D.
Note:
We need to keep in mind that if any two lines are parallel to each other, then the value of their slopes will be equal. However, if any two lines are perpendicular to each other, then the value of the product of their slopes will be equal to \[ - 1\]. We might get confused between abscissa and ordinate and find the ordinate of the point. In a Cartesian plane, the coordinate of \[x\] axis is called abscissa, whereas the coordinate of \[y\] axis is called ordinate.
Formula used:
If \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] are two points, then the slope of the line joining the two points is given by \[{\rm{Slope}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].
Complete step by step solution:
Here we need to find the abscissa of the point on the curve at which the tangent is drawn.
The given curve is \[y = 3{x^2} - 11x - 15\].
Let the coordinates of the point at which the tangent is drawn be \[\left( {a,b} \right)\].
Now, we will draw a rough diagram to show the given information.
![seo images](https://www.vedantu.com/question-sets/6cb3fe87-d817-4a6b-b198-28f882075e155782517916199812519.png)
Now, we will find the slope of the tangent on the given curve. For that, we will first find the derivative of the given equation of the curve.
We will differentiate the equation of the given curve with respect to \[x\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {3{x^2} - 11x - 15} \right)\]
On differentiating each term, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 6x - 11\]
Now, to find the slope of the tangent, we will substitute the value of the coordinates of the point of the tangent in the above equation.
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {a,b} \right)}} = 6a - 11\]
As it is given that the line joining the two given points i.e. \[\left( {5,5} \right)\] and \[\left( {11,227} \right)\] which lies on the curve and the tangent are parallel to each other, then the slope of the line and that tangent will be equal.
Now, we will calculate the slope of the line joining the points \[\left( {5,5} \right)\] and \[\left( {11,227} \right)\].
Substituting \[{x_1} = 5,{y_1} = 5,{x_2} = 11\] and \[{y_2} = 227\] in the formula \[{\rm{Slope}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\], we get
\[{\rm{Slope}} = \dfrac{{227 - 5}}{{11 - 5}}\]
Subtracting the terms, we get
\[ \Rightarrow {\rm{Slope}} = \dfrac{{222}}{6}\]
Dividing 222 by 6, we get
\[ \Rightarrow {\rm{Slope}} = 37\]
Now, we will equate the slope of the line and the slope of the tangent. Therefore, we get
\[6a - 11 = 37\]
Adding 11 on both sides, we get
\[ \Rightarrow 6a - 11 + 11 = 37 + 11\]
\[ \Rightarrow 6a = 48\]
Dividing both sides by 6, we get
\[ \Rightarrow \dfrac{{6a}}{6} = \dfrac{{48}}{6}\]
\[ \Rightarrow a = 8\]
Hence, the required abscissa of the point is equal to 8.
Therefore, the correct answer is option D.
Note:
We need to keep in mind that if any two lines are parallel to each other, then the value of their slopes will be equal. However, if any two lines are perpendicular to each other, then the value of the product of their slopes will be equal to \[ - 1\]. We might get confused between abscissa and ordinate and find the ordinate of the point. In a Cartesian plane, the coordinate of \[x\] axis is called abscissa, whereas the coordinate of \[y\] axis is called ordinate.
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