Tangents And Normals

Step 1: Find slope of the line joining the points: The curve is: Let the point of tangency be on the curve. We are told the line with slope -1 is tangent to the curve, so: Also, point lies on curve: Now since line is tangent to curve and also passes through both (0, 3) and (5, -2), every point on the curve and tangent must satisfy: So set: Max value of C occurs at vertex At :

Given the curve , and the line passes through and is tangent.
Let the point of tangency be . Then:
- Slope of tangent to curve at :
- Line joining to : slope =
Equating both slopes: Substitute into Only is positive and makes sense for a tangent from origin above the curve.

The curve has the property that the derivative at any point is: Suppose line is normal: slope =
So

We are given the curve and the point from which the tangents are drawn. The points of contact of the tangents with the curve are and . Step 1: Equation of the tangent to the curve The equation of the tangent to the curve at any point is given by: where is the derivative of . Differentiating: Substitute and into the tangent equation: Step 2: Finding the points of contact The tangents from the origin will satisfy the equation of the tangent. We use the condition that the distance from the origin to the tangent line is zero. After solving, we find that . Thus, the correct answer is option (3), .

Step 1: Set up the condition for the tangent.
The curve is , and is the tangent at for all . This means: - The point lies on the curve: - The line is the tangent, so the slope of the curve at must equal the slope of the line , which is 1, and the point must satisfy the line’s equation: - Compute the derivative of the curve to find the slope: Use the product rule: At , the slope must be 1:
Step 2: Solve the equations.
From (2), . Substitute into (1): Equation (3) must hold for all , so . This must be independent of . Test possible values for . Assume is constant for all , which implies the coefficient of must be zero: But if , from (4): This holds, but check (3): This satisfies the slope condition. So, , . Thus, the point is , and: This matches option (3).
Step 3: Verify with the point .
The curve at : , so is on the curve. The line at : , so it passes through . The slope condition already holds for all , confirming is the fixed point. Final Answer:

Given the slope of the tangent at any point is: At , substitute: So the slope of the tangent is The slope of the normal is the negative reciprocal: Using point-slope form for the line: Multiply both sides by 2:

Step 1: Verify points and lie on the curve.
The curve is .
For : , which matches.
For : , which matches.
Both points are on the curve.
Step 2: Find the slope of the tangent at .
The curve is . The derivative (slope of the tangent) is: At :
Step 3: Determine the slope of chord and set it equal to the tangent’s slope.
Let point be . Since is on the curve, , so .
The slope of chord between and is: The tangent at has slope , and this must equal the slope of : Thus, . So, .
Step 4: Verify lies on the curve and check options.
Point : , which matches. This corresponds to option (1).
Check other options:
: , on the curve, slope of : .
: , on the curve, slope: .
Option (1) is correct. Final Answer:

Step 1: Finding the equation of the line The equation of the line through and : Simplifying, Step 2: Condition for tangency The given curve is . For tangency, we equate slopes: Solving for , we get:

Step 1: Finding the derivative Step 2: Evaluating at Step 3: Using point-slope form Equation of the tangent: