Question

The tangents to the curve $y = {(x - 2)^2} - 1$ at its points of intersection with the line $x - y = 3$ , intersect at the point :
A) $\left( {\dfrac{{ - 5}}{2}, - 1} \right)$
B) $\left( {\dfrac{{ - 5}}{2},1} \right)$
C) $\left( {\dfrac{5}{2}, - 1} \right)$
D) $\left( {\dfrac{5}{2},1} \right)$

Answer 1

Hint: First we find the point of intersection of the two given curves by solving these equations. After finding intersection points, we get the slope of the tangent. Slope of the tangent is given by \[\dfrac{{dy}}{{dx}}\]. We get two equations of tangent. After solving these equations, finally we get the required intersection of points.

Formula used: First we calculate two points of intersection by solving two linear equations. Formula for the equation of tangent in two points is used in question. The formula of a straight line passing through two points is given by \[y - {y_1} = m\left( {x - {x_1}} \right)\]. Where slope \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].

Complete step-by-step answer:
First we have to find the point of intersection of the two given curves:
$y = {(x - 2)^2} - 1 ……………….(1)$
$x - y = 3 ……………….(2)$
Substituting the value of x from equation ( 2) in equation ( 1) , we get :
$
  y = {(3 + y - 2)^2} - 1 \\
  y = {(1 + y)^2} - 1 \\
  y = 1 + {y^2} + 2y - 1 \\
  {y^2} + y = 0 \\
  y(y + 1) = 0
 $
 $y = 0$ or \[y = - 1\]
Substituting these values of y in equation (2) we get: \[x = - 3\] or \[x = 2\].
Thus points of intersection of the two curves are\[\left( {3{\text{ }},{\text{ }}0} \right){\text{ }}and{\text{ }}\left( {2{\text{ }},{\text{ }} - 1} \right)\].
Now we must find the slope of tangents using differentiation.
\[
  y = {(x - 2)^2} - 1 \\
  y = {x^2} + 4 - 4x - 1 \\
  y = {x^2} - 4x + 3 \\
  \therefore \dfrac{{dy}}{{dx}} = 2x - 4
 \]
Slope of the tangent at \[(3,0)\] is \[2\] .
Equation of tangent at \[(3,0)\] is :
\[y = 2x - 6 ……………….(3)\]
Slope of tangent at \[\left( {{\text{ }}2{\text{ }},{\text{ }} - {\text{ }}1} \right)\] is 0.
Equation of tangent at \[\left( {{\text{ }}2{\text{ }},{\text{ }} - {\text{ }}1} \right)\] is:
\[y = - 1 ………………..(4)\]
Now the point of intersection of the lines represented by ( 3) and ( 4) is \[\left( {{\text{ }}5/2{\text{ }},{\text{ }} - {\text{ }}1} \right)\].

Hence (C) is the correct option.

Note: Slope of tangent to a curve \[y = f(x)\] is given by \[\dfrac{{dy}}{{dx}}\]. Students first arrange the equation of tangent in the linear equation as \[ax + by + c = 0\]. Then find differentiation with respect to x. Hence, we get\[\dfrac{{dy}}{{dx}}\] .
Any line which is parallel to the x – axis has slope equal to zero. And any line which is parallel to the y – axis has slope equal to infinity.
Also, the equation of a line passing through the point \[({x_1},{y_1})\] and having slope m , is given by : \[y - {y_1} = m(x - {x_1})\].