Conic Sections

Step 1: Understanding the Parabola and Given Points
We are given the parabola . The equation of the parabola represents a standard parabola that opens to the right with the vertex at the origin .
The focus of the parabola, denoted by , is at , since the general formula for a parabola of the form gives the focus at . Here, , so the focus is at .
We are also given the point and two distinct points and on the parabola such that the lines and are tangents to the parabola at and . We need to determine which of the following statements is true:

• (A) The triangle is a right-angled triangle
• (B) The triangle is a right-angled triangle
• (D) lies on the line joining and

Step 2: Statement (A) - The triangle is a right-angled triangle
To determine if the triangle is a right-angled triangle, we need to check if the angle is . We know that the tangent at any point on the parabola makes an angle of with the line joining the point of tangency and the focus. Hence, since is tangent to the parabola at , and the focus is at , the angle is indeed a right angle.
Therefore, statement (A) is TRUE.
Step 3: Statement (B) - The triangle is a right-angled triangle
For triangle to be right-angled, one of its angles must be . The points and are distinct points on the parabola, and the lines and are tangents to the parabola at these points. Since the tangents at two distinct points on the parabola are perpendicular to the line joining those points with the focus, the angle between and is . This makes triangle a right-angled triangle.
Therefore, statement (B) is TRUE.
Step 4: Statement (D) - lies on the line joining and
This statement is related to a well-known property of parabolas: the two tangents drawn from any external point to a parabola always meet at the same angle at the focus. Since and are points of tangency, the line joining them will pass through the focus . This is because the tangents at and must be symmetric with respect to the line joining and , and the focus lies on this line.
Therefore, statement (D) is TRUE.
Step 5: Conclusion
The correct statements are (A), (B), and (D). Thus, the answer is:

TRUE for all statements: (A), (B), (D).

Let and be and , respectively, and is given as .

Slope of is , and the slope of is infinity.

Equation of :

Equation of and :

Given: The equation , where and are distances, and the relationship involving the variable , , and .

Step 1: Equation 1

The first equation is:

Step 2: Simplification of the equation

Next, we solve the following equation for and :

Step 3: Relationship between and

We are given:

Step 4: Calculating

We now calculate using the formula:

Step 5: Applying the relation between and

Now, using the equation , we get the following relationship:

Step 6: Solving for and

We substitute the values:

From this, we find:

Conclusion

Therefore, the final result is .

Final Answer: The value of is 7, as required by the problem.

Equation of normal at P(am², –2am) is y = mx – 2am – am³
⇒ Area of ∆PFQ = 1/2(a + am²) x 2am = 120
a²m(1 + m²) = 120 Pair (a, m) ≡ (2, 3) satisfies above equation.

Given :
, Tangent :
P : y² = 12x, Tangent :
Now, for common tangent :

⇒ m = ±1
The equations of the common tangents are y = x + 3 and y = -x - 3. The points of contact for the parabola are:

A₁ ≡ (3, 6), A₄(3 - 6)
Now, Let suppose A₂(x₁, y₁)
⇒ tangent to E :
A₃ is mirror image of A₂ in x-axis ⇒ A₃(–2, –1)

The intersection point of T1 = 0 and T2 = 0 is at (-3, 0).
Area of quadrilateral A₁A₂A₃A₄ :
square units.
So, the correct options are (A) and (C).

Equation of the Normal and Intersection Points

The equation of the normal to is:

With slope , we have .

Substituting :

Points of Intersection A and B:

Length of AB:

Simplifying:

Given:

Solving:

Thus, .