Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2010 | Oct-Nov | (P2-9709/23) | Q#8

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  Question

The equation of a curve is x2 + 2xy y2 + 8 = 0.

     i.       Show that the tangent to the curve at the point (-2,2) is parallel to the x-axis.

   ii.       Find the equation of the tangent to the curve at the other point on the curve for which x=-2,  giving your answer in the form y = mx + c.

Solution

     i.
 

We are given equation of the curve as;

If two lines are parallel to each other, then their slopes  and  are equal;

Hence, if tangent to the curve at point (-2,2) is parallel to the x-axis then slope of that tangent must  be equal to slope of the x-axis ie ZERO.

Therefore, we need to show that slope of the tangent to the curve at point (2,-2) is ZERO.

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the  same point;

Therefore, if we can find slope of the curve at the point (-2,2) the same will be slope of tangent at  this point to the curve.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to  is:

We are required to find .

To find from an implicit equation, differentiate each term with respect to , using the chain rule to  differentiate any function  as .

Rule for differentiation of  is:

We differentiate each term of the equation, one by one, with respect to x applying following rules. 

Rule for differentiation of  is:

If  and  are functions of , and if , then;

Rule for differentiation of  is:

Now we can combines derivatives of all terms of the equation as;

Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.

Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve;

We can substitute coordinates of the point (2,-2) to find the gradient at this point.

Therefore, gradient of the curve at point (-2,2), and hence that of tangent to the curve at this point,  is ZERO.

It is evident that slope of tangent to the curve at point (-2,2) is equal to slope of the x-axis ie both  are parallel.

 

   ii.
 

Next we are required to find the equation of the tangent to the curve at point (-2,y).

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

Therefore, we need to find the y-coordinate of the point (-2,y).

Corresponding values of y coordinate can be found by substituting value of x in equation of the line  or the curve.

We are given equation of the curve as;

We substitute x=-2 in the given equation;

Now we have two options.

Therefore, coordinates of this other than (-2,2) point are (-2,-6).

Next we need slope of the tangent to the curve at point (-2,-6).

The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the  same point;

We have already found in (i) expression for gradient of the curve as;

We need gradient of curve at point (-2,-6).

Therefore slope of the curve at point (-2,-6) is;

Hence, slope of the tangent to the curve at this point is;

Now we can write equation of the tangent as follows.

Point-Slope form of the equation of the line is;

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