A triangle ABC with area 5a^{2} is inscribed | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the coordinates of the endpoints of the focal chord $BC$ be $B(at_{1}^{2}, 2at_{1})$ and $C(at_{2}^{2}, 2at_{2})$.Since $BC$ is a focal chord,

Vedclass · Accepted Answer

$25a$

Vedclass · Answer

(D) Let the coordinates of the endpoints of the focal chord $BC$ be $B(at_{1}^{2}, 2at_{1})$ and $C(at_{2}^{2}, 2at_{2})$.Since $BC$ is a focal chord,$t_{1}t_{2} = -1$.The area of $\Delta ABC$ with vertices $A(0, 0)$,$B(at_{1}^{2}, 2at_{1})$,and $C(at_{2}^{2}, 2at_{2})$ is given by:$	ext{Area} = \frac{1}{2} |x_{A}(y_{B} - y_{C}) + x_{B}(y_{C} - y_{A}) + x_{C}(y_{A} - y_{B})|$$	ext{Area} = \frac{1}{2} |0 + at_{1}^{2}(2at_{2} - 0) + at_{2}^{2}(0 - 2at_{1})| = \frac{1}{2} |2a^{2}t_{1}^{2}t_{2} - 2a^{2}t_{1}t_{2}^{2}|$$	ext{Area} = a^{2} |t_{1}t_{2}(t_{1} - t_{2})| = a^{2} |-1(t_{1} - t_{2})| = a^{2} |t_{1} - t_{2}|$.Given that the area is $5a^{2}$,we have $a^{2} |t_{1} - t_{2}| = 5a^{2}$,which implies $|t_{1} - t_{2}| = 5$.The length of the focal chord is $L = a(t_{1} - t_{2})^{2}$.Since $t_{2} = -1/t_{1}$,we have $t_{1} - t_{2} = t_{1} + 1/t_{1}$.Thus,$(t_{1} - t_{2})^{2} = (t_{1} + 1/t_{1})^{2} = (t_{1} - 1/t_{1})^{2} + 4 = (t_{1} - t_{2})^{2} + 4 = 5^{2} + 4 = 29$.Wait,the length of the focal chord is $a(t_{1} - t_{2})^{2} = a(t_{1} + 1/t_{1})^{2}$.Using $|t_{1} - t_{2}| = 5$,we have $(t_{1} - t_{2})^{2} = 25$.Since $t_{1}t_{2} = -1$,$(t_{1} + t_{2})^{2} = (t_{1} - t_{2})^{2} + 4t_{1}t_{2} = 25 - 4 = 21$.The length of the focal chord $BC$ is $a(t_{1} - t_{2})\sqrt{(t_{1} + t_{2})^{2}} = a(5)\sqrt{21}$ is not correct.The length of the focal chord is $a(t_{1} - t_{2})^{2} = a(t_{1} + 1/t_{1})^{2} = a(t_{1}^{2} + 1/t_{1}^{2} + 2) = a((t_{1} - 1/t_{1})^{2} + 4) = a(25 + 4) = 29a$.Re-evaluating: The length of the focal chord is $a(t_{1} + 1/t_{1})^{2}$. Given $|t_{1} - t_{2}| = 5$,where $t_{2} = -1/t_{1}$,then $|t_{1} + 1/t_{1}| = 5$. Thus,length $= a(5)^{2} = 25a$.

$A$ triangle $ABC$ with area $5a^{2}$ is inscribed in the parabola $y^{2} = 4ax$,where the vertex $A$ is at the vertex of the parabola and $BC$ is a focal chord. Find the length of the focal chord.

Similar Questions

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $y^{2} = -8x$.

The point of intersection of the normals to the parabola $y^2 = 4x$ at the ends of its latus rectum is

$A$ chord is drawn through the focus of the parabola $y^2 = 6x$ such that its distance from the vertex of this parabola is $\frac{\sqrt{5}}{2}$. Then,its slope can be:

What is the area of the triangle formed by the vertex of the parabola $x^2 = 12y$ and the endpoints of its latus rectum (in square units)?

Which one of the following curves cuts the parabola $y^2 = 4ax$ at right angles?

Vedclass Test Series

Exam Paper Generator

Online Exam Module