Tangents And Normals

The correct answer is 45

so f(x) is decreasing in and increasing in

Tangent at

It is passing through (4, 3)


So tangent may be

But passes through (–2, 0) so rejected.
Equation of Normal



α + β = 9 + 36
= 45

The correct answer is (D) : S=N




So line always touches the given curve.

The correct answer is (B) :



∴ Solution of D.E. will be


∵ Curve passes through

The correct answer is (C) :


Centre of G=A




∴ area of trapezium:

Equation of the Curve

The equation of the curve is:

Step 1: Slope of the Normal Line

The normal line is parallel to the line , which has a slope:

The slope of the normal line is also . The normal slope is related to the derivative of the curve by:

Equating the slopes:

This simplifies to:

Step 2: Compute

The derivative of the curve is:

Differentiate each term:

We are given that , so substitute and simplify:

Subtract 90 from both sides:

Step 3: Solve the Equation

The equation:

has 4 real roots, corresponding to 4 points on the curve.

Conclusion

The number of points where the normal lines are parallel to is:

The correct answer is : 10

Step 1: The ellipse and the point of tangency.
The given ellipse is and the point of tangency is . One can verify that indeed lies on this ellipse.

Step 2: Finding the tangent and normal at .
The equation of the tangent to at the point is Hence at , the tangent is This line meets the -axis where , giving So the tangent intersects the -axis at . The slope of this tangent is found by rewriting Hence the slope is . The normal at is perpendicular to this tangent, so its slope is . Its equation going through is Substitute to find where it meets the -axis: so . Thus

Step 3: The circle with diameter .
The endpoints and . The midpoint is and the radius is . So has center and radius . Its equation:

Step 4: Intersection of the line with .
Substitute into the circle: Hence or So the points of intersection are

Step 5: Tangents at and to the circle, and their intersection .
The tangent to the circle at a point on is: At , the tangent is At , the tangent is We solve (T ) and (T ) simultaneously: Add these two equations: Now substitute back (e.g. into T ): This becomes Hence the tangents intersect at . Thus and

Step 6: Calculate .
Hence Hence the final result given is

Given: ,

Step 1: Equation of the tangent to the parabola: The equation of the tangent to the parabola is:

... (1)

This tangent also touches the circle .

Step 2: Condition for tangency: The perpendicular distance from the center of the circle (4,0) to the tangent is equal to the radius of the circle:

Step 3: Solve for m: Square both sides:

Multiply through by :

Solve for distinct :

Step 4: Point of contact of the parabola: For , substitute into :

Step 5: Calculate : The distance between P and Q is given by:

Final Answer:

Shaded Area Calculation

The equation of the curve is given as:

The tangent at the point is:

Step 1: Solve the Integral

We compute the integral of the curve from to :

Evaluating the integral:

Step 2: Compute the Areas of the Triangles

1. Area of Triangle

The area of is:

2. Area of Triangle

The vertices of are .

We compute the area using the formula for the area of a triangle with given vertices:

3. Area of Triangle

The vertices of are .

We compute the area similarly:

Simplifying the expression:

Step 3: Combine the Results

Now, we combine all the computed areas:

To simplify, we find the common denominator (120):

Final Answer

The shaded area is .

Solution:

The equation of the tangent to the hyperbola is given by:

,

where and .

For the point , substitute :

.

Simplifying:

.

Squaring both sides:

.

Expanding:

.

Combining terms:

.

Factoring:

.

Thus, and .

The equation of the tangent with is:

,

which intercepts the x-axis at .

The equation of the tangent with is:

,

which intercepts the x-axis at .

Intersection Point Q:

Solve the system of equations:

Equating the y-values:

Substitute into :

So, .

Distance PQ:

.

Calculate the ratio:

.

There seems to be an error in the original problem or the desired answer, as the final value is 0.

We are given the equation of the circle: First, rewrite the equation in standard form by completing the square.

Step 1: Completing the square. For , we have: For , we have: Substitute these into the original equation: Simplifying: Thus, the equation of the circle is: which shows that the center of the circle is and the radius is .

Step 2: Using the formula for the length of the tangent from an external point. The length of the tangent from a point to a circle with center and radius is given by: For point , the tangent length is: Simplifying: Thus, the length of the tangent is . The correct answer is option (3).

Step 1: Finding the slope of the reflected ray.
The light ray emits from the origin and makes an angle of with the positive x-axis. The slope of the reflected ray is: Step 2: Finding the equation of the reflected ray. The equation of the reflected ray will be in the form: where is the slope and is the y-intercept. Since the ray passes through the origin, . Therefore, the equation of the reflected ray is: Step 3: Finding the intersection point with the line .
The line intersects the reflected ray at the point where the equation satisfies the equation . Substituting into the line equation: Step 4: Finding the abscissa of .
The abscissa of is found by substituting into the equation of the reflected ray . The value of gives the abscissa of the point where the ray intersects the x-axis. Rationalizing the denominator, we get: Thus, the abscissa of is , and the correct answer is option (2).

Parabola Tangent and Intersection Problem

The given curve is . Completing the square, we get

This is a parabola with vertex (-1, 2).

The equation of the tangent to the parabola at is given by

So, the equation of the tangent at is

This tangent meets the y-axis at A. So, the coordinates of A are (0, 2).

The line passing through and parallel to is given by

So the equation of the line is .

Given there appears to be an error stating that parabola is when in fact the original is , let us assume that another curve intersects with the previous tangential point to be solved accordingly.

To proceed, we will continue what would of happened: This line intersects the parabola . Substituting in , we have

So, or .

If , . If , .

The points of intersection are (4, 4) and (16, 8).

We are given that B lies on the line .

For (4, 4): . For (16, 8): . So, B is (16, 8).

Now, A is (0, 2) and B is (16, 8).

Since there may be a mistake from the original problem, will make notes to be fixed here

Conclusion:

To solve this problem, we need to determine the tangent of the angle between the lines and which trisect the portion of the line that lies in the first quadrant.

Step-by-step Solution

1. First, find the intercepts of the line in the first quadrant.
   • When , (x-intercept)
   • When , (y-intercept)
2. Determine the points that trisect this line segment.
   • Using section formula, the points that divide the segment in the ratio 1:2 and 2:1 are:
     • First point (Divide in 1:2 ratio):
     • Second point (Divide in 2:1 ratio):
3. Find the equations of the lines and that pass through these points from the origin.
   • Line through :
     Equation: , where .
   • Line through :
     Equation: , where .
4. Find the tangent of the angle between these lines.
   • The formula for the tangent of the angle between two lines and is:
   • For and , substitute into the formula:

Therefore, the tangent of the angle between the lines and is .