Take the Tangent to the Curve at Q Ot Intersect the Curve Again

Question

The signal A(2,2) lies on the curve .

i. Find the equation of the tangent to the bend at A.

The normal to the curve at A intersects the curve again at B.

2. Discover the coordinates of B.

The tangents at A and B intersect each other at C.

iii. Find the coordinates of C.

Solution

i.

We are required to discover the equation of the tangent to the curve at A(2,2).

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

We already take coordinates of a point on the tangent A(ii,ii). Even so, we need to find slope of  the tangent to write its equation.

The slope of a curve  at a particular betoken is equal to the slope of the tangent to the curve at the  aforementioned betoken;

Therefore, we can detect the slope of tangent from slope of the curve at bespeak A(2,2).

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

We are given equation of the curve every bit;

Therefore;

Rule for differentiation of  is:

Dominion for differentiation of  is:

Dominion for differentiation of  is:

This gives the expression for gradient of the bend.

At present nosotros need to find gradient of the curve at point A(2,2).

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

Therefore, substituting  in expression of the gradient of the bend;

Hence slope of the curve at point A is , and therefore, gradient of the tangent to the curve at  point A;

Now we tin can write equation of the tangent as follows.

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

ii.

We are required to find the coordinates of point B which is intersection bespeak of the curve and the  normal to the curve at point A(2,2).

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e.  coordinates of that point have aforementioned values on both lines (or on the line and the curve).  Therefore, we tin equate  coordinates of both lines i.e. equate equations of both the lines (or the  line and the bend).

Therefore, we need equations of both the curve and the normal to the curve at point A(2,2).

Nosotros are already given equation of the curve as;

Now nosotros need to find equation of normal to the curve at point A(two,two).

To find the equation of the line either nosotros need coordinates of the two points on the line (2-Indicate  form of Equation of Line) or coordinates of one point on the line and slope of the line (Signal-Slope  form of Equation of Line).

Nosotros already have coordinates of a signal on the normal A(two,2).
However, nosotros need to notice gradient of the normal to write its equation.

If a line  is normal to the curve , and then production of their slopes  and  at that point (where line is normal to the bend) is;

Therefore, we can detect the slope of normal from gradient of the curve at indicate A(ii,ii).

We accept already found in (i) slope of the curve at point A as , and therefore, gradient of the  normal to the curve at point A;

At present we tin write equation of the normal as follows.

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

Now when we have equations of both the bend and the normal to the bend at bespeak A(2,2), nosotros can  equate these two equations to find the coordinates of point of intersection.

Now we accept ii options.

Two values of 10 betoken that in that location are ii intersection points.

It is axiomatic that  represents point A(2,two) where line is normal to the bend and represents indicate B where normal over again meets the bend.

With ten-coordinate of point of intersection of two lines (or line and the curve) at hand, nosotros can detect the  y-coordinate of the point of intersection of two lines (or line and the curve) past substituting value  of x-coordinate of the point of intersection in whatever of the ii equations;

Nosotros choose equation of normal;

Therefore, we substitute ;

Hence, coordinates of point B are .

three.

We are required to observe the coordinates of betoken C which is intersection betoken of the tangent to the  bend at point A(two,2) and tangent to the curve at point B .

If two lines (or a line and a bend) intersect each other at a signal so that point lies on both lines i.e.  coordinates of that betoken take same values on both lines (or on the line and the curve).  Therefore, we can equate  coordinates of both lines i.e. equate equations of both the lines (or the  line and the bend).

Therefore, nosotros demand equations of both the tangents to the bend at point A and B.

We already have found equation of tangent to the curve at point A(2,ii) as;

However, we demand to find equation of tangent to the bend at point B .

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

We already accept coordinates of a bespeak on the tangent B . However, nosotros need to find slope of  the tangent to write its equation.

The slope of a bend  at a particular indicate is equal to the slope of the tangent to the curve at the  aforementioned point;

Therefore, we can find the slope of tangent from gradient of the curve at point B .

We take already found in (i) expression for the slope of the curve;

Now we demand to find gradient of the curve at signal B .

Gradient (slope)  of the bend  at a item  betoken  can exist institute by substituting x- coordinates of that point in the expression for gradient of the curve;

Therefore, substituting  in expression of the slope of the curve;

Hence slope of the curve at betoken B is , and therefore, slope of the tangent to the curve  at point B;

Now we tin can write equation of the tangent as follows.

Indicate-Gradient course of the equation of the line is;

Equation of tangent at A is;

Equation of tangent at B is;

Equating both equations;

Unmarried value of x indicates that there is only ane intersection point.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand, we tin can discover the  y-coordinate of the bespeak of intersection of two lines (or line and the curve) by substituting value  of ten-coordinate of the point of intersection in whatsoever of the ii equations;

We cull equation of tangent;

Therefore, we substitute ;

Hence, coordinates of point C are .

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