Straight Lines

ax² + 2hxy + by² =0
m₁ + m₂ = -2h/b
m₁m₂ =
Given:
m₁ = 2m₂
put m₁ = 2m₂
2m₂ + m₂ =
3m₂ =
m₂ =
and 2m₂ =
m² =
Now, =
=
8h² = 9ab
Therefore the correct option is (A) 8h² = 9ab

Step 1: Find the Gradient of the Given Line
The given line is . Rearranging to slope-intercept form , we get: Therefore, the gradient of the line is .

Step 2: Differentiate the Curve Implicitly
The given curve is . Differentiating implicitly with respect to , we get: Solving for , we find: This is the gradient of the tangent to the curve.

Step 3: Gradient of the Normal
The normal to the curve is perpendicular to the tangent. Thus, the gradient of the normal is the negative reciprocal of the tangent's gradient:

Step 4: Use the Parallel Condition
The normal is parallel to the given line, which has a gradient of . Therefore, the gradient of the normal must also be , leading to: This gives:

Step 5: Find the Point of Intersection
Substitute into the equation of the curve: Simplifying: So:

Final Answer: The points of intersection are .

Step 1: Start with the equation: Expanding both sides: Simplifying:

Step 2: Substitute into the equation: Simplifying: Solving for :

Step 3: Apply the condition for perpendicular lines: Solving for :

Step 4: The smallest positive value of is:

Final Answer: The smallest positive value of is .

The points (1, 3) and (5, 1) are opposite vertices of a diagonal of a rectangle. We are given that the other two vertices lie on the line , and we need to find one of the vertices on the other diagonal.

Step 1: Midpoint of the diagonal:
The midpoint of the diagonal connecting the points (1, 3) and (5, 1) can be found by averaging the coordinates of these two points. The midpoint formula is: Substituting the coordinates of the given points: So, the midpoint of the diagonal is .

Step 2: Midpoint of the other diagonal:
The opposite diagonal of the rectangle will also have its midpoint at the same point , as the diagonals of a rectangle bisect each other. If the other two vertices of the rectangle are and , then the midpoint of this diagonal is: Therefore, we have the following system of equations: This simplifies to:

Step 3: Use the line equation:
We are told that the other two vertices lie on the line . Let’s substitute the coordinates of the points and into the equation of the line: Now substitute these into the equation for : Simplifying: Using the equation from earlier: Therefore, the equation of the line becomes .

Step 4: Find the points on the line:
Now that we know the equation of the line is , we can substitute the values of and into this equation. We know that . Let’s solve for the coordinates of the points. Assume and (since their sum must be 6). Substituting into the line equation: So, the points are and .

Final Answer:
One of the vertices on the other diagonal of the rectangle is . Therefore, the correct option is (C) : (2, 0).

Step 1: Find the slope of the line :
The given line is . We can rewrite it in slope-intercept form by solving for : Therefore, the slope of the line is .

Step 2: Find the slope of the curve :
The given curve is , which can be rewritten as: This is a hyperbola, and its slope at any point is given by the derivative of . Differentiating , we get: Therefore, the slope of the curve at any point on the curve is . However, for the purposes of this problem, we are concerned with the condition that the line is normal to the curve at some point. A line normal to the curve has a slope that is the negative reciprocal of the slope of the curve.

Step 3: Condition for normality:
For the line to be normal to the curve , the product of the slopes of the line and the curve at the point of intersection must be equal to . The slope of the line is , and the slope of the curve is . For the line to be normal to the curve, we require: Simplifying:

Step 4: Solve for the relationship between and :
Since cannot be zero, the above equation implies that: Therefore, we conclude that , or equivalently, and must be of the same sign (either both positive or both negative).

Conclusion:
The relationship between and is that and must either both be positive or both be negative for the line to be normal to the curve .