A curve is given by the equations x=sec^2 θand y=cotθ. If the tangent at P, where θ=(π)/(4) meet

A curve is given by the equations x=sec^2 θand y=cotθ. If...

Key Concepts

Distance Between Two Points

The distance between two points on a curve can be computed using the distance formula from coordinate geometry. This formula calculates the straight-line distance by taking the square root of the sum of the squares of the differences in the coordinates.

Tangent Lines to a Curve

A tangent line to a curve at a particular point is a straight line that just touches the curve without crossing it at that point. Its slope is equal to the derivative of the curve at that point, and it is fundamental in understanding the local linear behavior of the curve.

Intersection of a Curve and Its Tangent

Determining the points where a tangent line intersects the curve again involves solving the system of equations defined by the curve's parametric or Cartesian equations and the equation of the tangent line. This process is important in the study of curve properties and their geometrical relationships.

Parametric Equations

Parametric equations represent a curve where the coordinates are given as functions of a parameter. This approach allows one to express complex curves that might be difficult or even impossible to represent with a single Cartesian equation.

Parametric Differentiation

Parametric differentiation involves finding the derivatives of the coordinate functions with respect to the parameter. It is used to compute the slope of the tangent to a parametric curve by taking the ratio of dy/d? to dx/d?.

Transcript

0:00 Hello everyone.

00:02 From the question we are given that a curve is given by the equation x is equal to sec square theta and y is equals to chord theta.

00:11 If the tangent at p where theta equals to pi by 4 meets the curve again at q then we have to find the length of pq since we have x is equals to sec square theta and y is equals to cod theta.

00:24 So first differentiate with respect to theta then d x upon d theta is equal to 2 sec square theta 10 theta and d y by d theta is equal to minus cossack square theta so d y by d x is equal to d y by d theta divided by d x upon d theta this is equals to minus cossack square theta divided by 2 sac square theta 10 theta this is equals to minus 1 upon 2 cot square theta since we have theta is equal to pi by 4 so d y by d x at theta is equals to pi by 4 this is equals to minus 1 upon 2 so the point p is 2.

01:11 Now equation of the tangent at p is y minus 1 is equal to minus 1 upon 2 multiply by x minus 2 this implies 2y minus 2 is equals to minus x plus 2 this implies x plus 2y is equal to 4 this is the equation first now eliminating theta between x is equals to sec square theta and y is equals to cod theta then we get x minus 1 upon y square is equal to 1 this is the equation second by solving these two equations we get 4 minus 2y minus 1 upon y square is equal to 1 or we can write it as 4 y square minus 2 y cube minus 1 is equals to y square this implies 3y square minus 2 y cube minus 1 is equal to 0 this implies 2 y cube minus 3 y square plus 1 is equal to 0 this implies 2y cube minus 2 y square minus y square plus 1 plus y minus y minus 1 plus y minus minus y is equal to 0 add y and subtract y in this equation this implies 2 y square multiply by y minus 1 minus y multiply by y minus 1 minus 1 minus 1 minus y minus 1 is equals to 0 this implies y minus 1 multiply by 2 y square minus y minus 1 is equals to 0 if multiplication of y minus 1 and 2 y square minus 1 equals to 0 then we can write y minus 1 equals to 0 or 2 y square minus y minus 1 equals to 0 from here y is equals to 1 and from here y minus 1 multiply by 2y plus 1 is equal to 0 this implies y is equal to minus 1 upon 2 so y is equal to 1 comma minus 1 upon 2 so when y is equal to minus 1 upon 2 then x is equal to 4 minus 2y is equal to 4 plus 1 is equal to 5 so the point q is 5 5 minus 1 upon 2 since point p is 2 .1 and point q is 5 .5 minus 1 minus 1 upon 2...

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