The curve y=x^2-3x-3 crosses the x-axis at P and Q. The tangents to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS.
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,349 answers
starTop subjects are Math, Science, and Business
You need to remember that x intersections P and Q represent the roots of the equation `x^2-3x-3 = 0` , hence, you need to use the quadratic formula to find x coordinates of the points P and Q such that:
`x_(P,Q) = (3+-sqrt(9+12))/2 => x_(P,Q) = (3+-sqrt21)/2`
You need to find the equations of the tangent lines to the given curve, at the points P and Q such that:
`y - y_P = f'(x_P)(x - x_P)`
`y - y_Q = f'(x_Q)(x - x_Q)`
You need to evaluate `f'(x_P) ` and`f'(x_Q)` such that:
`f'(x_P) = (x^2_P-3x_P-3) => f'(x_P) = 2x_P - 3`
Substituting `(3+sqrt21)/2` for `x_P` yields:
`f'(x_P) = 2(3+sqrt21)/2 - 3 => f'(x_P) = 3+sqrt21-3`
`f'(x_P) = sqrt21`
You may write the equation of tangent line at the point P such that:
`y - 0 = sqrt21(x - (3+sqrt21)/2) => y = sqrt21(x - (3+sqrt21)/2)`
You need to evaluate `f'(x_Q)` such that:
`f'(x_Q) = 2x_Q - 3 => f'(x_Q) = -sqrt21`
You may write the equation of tangent line at the point Q such that:
`y - 0 = -sqrt21(x - (3-sqrt21)/2) => y = -sqrt21(x - (3-sqrt21)/2)`
The problem provides the information that the tangent lines intersect at R, hence, you need to solve the following system of equations to find R such that:
`{(y = sqrt21(x - (3+sqrt21)/2)),(y = -sqrt21(x - (3-sqrt21)/2)):}` =>`sqrt21(x - (3+sqrt21)/2) = -sqrt21(x - (3-sqrt21)/2) => x - (3+sqrt21)/2 = (3-sqrt21)/2 - x`
You need to move the terms that contain x to the left side such that:
`x + x = (3-sqrt21)/2+(3+sqrt21)/2 => 2x = 6/2 => 2x = 3 => x = 3/2`
`y =sqrt21(3/2 - (3+sqrt21)/2) => y = sqrt21*(-sqrt21)/2 = -21/2`
Hence, evaluating the coordinates of the point of intersection of tangents at P and Q, yields `R(3/2,-21/2).`
You need to find the equations of the normal lines at P and Q, hence, you need to remember that these normal lines are perpendicular to the tangent lines at P and Q.
You need to remember the equation that relates two perpendicular lines such that:
`m_1*m_P = -1`
`m_1` represents the slope of tangent line
`m_P ` represents the slope of normal line
Notice that the slope of tangent line at P is `m_1 = sqrt21` , hence, you may evaluate `m_P` such that:
`m_P = -1/sqrt21`
You need to use the point slope form of equation of line to write the equation of normal line at the point P, such that:
`y - y_P = m_P(x - x_P) => y = (-1/sqrt21)(x - (3+sqrt21)/2)`
Notice that the slope of tangent line at Q is `m_2 = -sqrt21` , hence, you may evaluate `m_Q` such that:
`m_Q = 1/sqrt21`
You need to use the point slope form of equation of line to write the equation of normal line at the point Q, such that:
`y - y_Q = m_Q(x - x_Q) => y = (1/sqrt21)(x - (3-sqrt21)/2)`
The problem provides the information that the tangent lines intersect at S, hence, you need to solve the following system of equations to find R such that:
`{(y= (-1/sqrt21)(x - (3+sqrt21)/2)),(y = (1/sqrt21)(x - (3-sqrt21)/2)):}` `=> (-1/sqrt21)(x - (3+sqrt21)/2)= (1/sqrt21)(x - (3-sqrt21)/2)` => `-x+ (3+sqrt21)/2 = x - (3-sqrt21)/2` => `2x = 6/2 => 2x = 3 => x = 3/2`
`y = (1/sqrt21)(3/2 - 3/2 + sqrt21/2) => y = 1/2`
Hence, evaluating the point of intersection of normal lines yields `S(3/2,1/2).`
You need to use the distance formula to evaluate the distance RS such that:
`RS = sqrt((x_S - x_R)^2 + (y_S - y_R)^2)`
`RS = sqrt((3/2 - 3/2)^2 + (1/2+ 21/2)^2) `
`RS = sqrt((22/2)^2) => RS = sqrt(11^2) = 11`
Hence, evaluating the distance RS, under the given conditions, yields `RS = 11` .
Related Questions
- The curve y=2x²-3x-4 crosses the x-axis at P and Q.The tangents to the curve at P and Q meet at...
- 1 Educator Answer
- The curve y=x^2-7x+6 crosses the x-axis at P and Q. The tangents to the curve P and Q meet at R....
- 1 Educator Answer
- At what point on the curve y = x - 3x^2 + 2 is the tangent parallel to the x-axis.
- 1 Educator Answer
- `x=t^3-6t , y=t^2` Find the equations of the tangent lines at the point where the curve...
- 1 Educator Answer
- find equations of the tangent line and normal line to the given curve at the specified point??...
- 1 Educator Answer
According to my book, that's not right...
y = x^2 - 3x -3
solve the quadratic to find the points at which y crosses the x-axis i.e.
x = (3 + (21)^0.5)/2 and x = (3 - (21)^0.5)/2
y' = 2x - 3
So the gradients of y at P and Q are -(21)^0.5 and (21)^0.5 respectively.
The normals will then have gradients of 21^-0.5 and -21^-0.5 respectively
By symmetry we can deduce that the tangents will meet where x = 3/2 and so will the normals
The gradient of the tangent from P is -21^0.5 and goes a distance of 21^0.5 /2 so the co-ordinates of R=(3/2,-21/2).
The gradient of the normal from P is 21^-0.5 and goes a distance of 21^0.5 /2 so the co-ordinates of S=(3/2, 1/2)
Hence the distance RS = 1/2 + 21/2 = 11
Student Answers