Tangents And Normals

Step 1: Differentiate implicitly: 4x-6y(dy)/(dx)=0
Step 2: (dy)/(dx)=(2x)/(3y)
Step 3: At (3,2), (dy)/(dx)=-1

Step 1: Differentiate implicitly. (dy)/(dx)=eˣ-1
Step 2: Vertical tangent condition. (dy)/(dx)→∞ eˣ-1=0 x=0
Step 3: Corresponding y. y=1

Step 1: Use the slope of the normal.
The slope of the normal is the negative reciprocal of the slope of the tangent. The slope of the normal is given by , so the slope of the tangent is 4/3. Therefore, .
Thus, the correct answer is (3).

Step 1: Find the derivative of the curve.
The derivative of the curve is given by the formula , which gives the slope of the tangent.
Step 2: Use the tangent slope condition.
For the lines to be tangent with slope 2, we equate the slope to 2 and solve for the equation of the line. Final Answer:

Equation of pair of tangents is given by
or ,

Step 1:
Step 2: Minimise derivative:

Step 1: Tangent condition from origin:
Step 2: Eliminating parameter does not yield any of the given relations.

Step 1: Slope of curve:
Step 2: Angle between tangent and x-axis is :
Step 3:

Step 1: Tangent parallel to x-axis ⟹ f'(x)=0.
Step 2: Differentiate: f'(x)=\frac12√(x)(7x-6)+7√(x)
Step 3: Set to zero and simplify: frac7x-6+14x2√(x)=0 ⟹ 21x-6=0 ⟹ x=(2)/(7)

(dy)/(dx)=eˣ(cos x-sin x) The slope is minimum when cos x-sin x is minimum, which occurs at x=π

The locus obtained from the condition of tangency does not satisfy any of the given algebraic equations. Hence, the correct answer is None of these.

Tangent parallel to x-axis ⟹ f'(x)=0. Differentiating: f'(x)=\frac7x-62√(x)+7√(x) Setting f'(x)=0: 7x-6+14x=0 ⟹ 21x=6 ⟹ x=(6)/(7)

To find the equation of the tangent to the curve at the point where , we need to follow these steps:

Therefore, the equation of the tangent line is , which matches the second option.

Step 1: Understanding the Concept
A normal to a curve at a point is a line perpendicular to the tangent at that point. To find its equation under a specific geometric constraint (perpendicular to another line), we need to determine the slope and the point of contact.

Step 2: Detailed Explanation
1. Differentiation (B): We differentiate the equation of the parabola ( ) to find the slope of the tangent ( ). The slope of the normal is then . 2. Slope Comparison (A): The problem states the normal is perpendicular to a "given line." If the given line has slope , then the slope of our normal must be . We compare this required slope to the derivative-based slope to find the specific point on the parabola. 3. Synthesis: You cannot solve the problem without finding the derivative (to relate the point to the slope) and comparing slopes (to apply the perpendicularity condition).

Step 3: Final Answer
The process requires both slope comparison and differentiation.