the tangent yo the curve y=x^2 +5x -2 @ the point (1,4)intersect the normal to the same curve @ the point (-3,-8) at the point P.Find the coordinates of point P.[ans: -1/3,-16/3]
just give me some hint to calculate this solution.
dy/dx = 2x + 5
slope of tangent at (1,4) = 2(1) +5 = 7
at (1,4) ...
4 = 7(1) + b
b= -3
tangent equation at (1,4) is y = 7x - 3
at (-3,-8)
slope of tangent is 2(-3) + 5 = -1
so slope of normal is +1
equation of normal is y = x + b
at (-3,-8) , -8 = -3 + b
b = -5
equation of normal is y = x - 5
where do they intersect ?
when 7x - 3 = x - 5
6x = -2
x = -1/3
back into y = x-5 , (or the other equation, makes no differencen)
y = - 1/3 - 5 = -1/3 - 15/3 = -16/3
so the two straight lines intersect at (-1/3, -16/3)
To find the coordinates of point P where the tangent and normal to the curve intersect, you'll need to follow these steps:
1. Find the derivative of the curve equation y = x^2 + 5x - 2 to get the slope of the tangent line at any given point on the curve.
2. Substitute x = 1 into the derivative to find the slope of the tangent line at the point (1, 4).
3. The slope of the tangent line is also the negative reciprocal of the slope of the normal line (perpendicular to the tangent line).
4. Find the equation of the line perpendicular to the tangent line at the point (-3, -8) using the negative reciprocal slope found in step 3. This line will be in the form y = mx + c, where m is the slope and c is the y-intercept.
5. Substitute x = -3 into the equation obtained in step 4 to find the y-coordinate of point P.
6. Solve the system of equations formed by the curve equation and the equation of the normal line to find the x-coordinate of point P.
7. The coordinates of point P will be (-1/3, -16/3).
This process uses calculus concepts to find the slope of the tangent line and the negative reciprocal slope for the normal line. Then, by solving the system of equations, we can find the intersection point P.