ParabolaMedium
Question
Consider a branch of the hyperbola x2 - 2y2 - 2 √2x - 4√2y - 6 = 0 with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is
Options
A.
B.
C.
D.
Solution
Hyperbola is 
a = 2, b = √2
e =
Area =
a(e - 1) × 
⇒ Area =
.
a = 2, b = √2
e =
Area =
a(e - 1) × ⇒ Area =
.Create a free account to view solution
View Solution FreeMore Parabola Questions
The range of $\alpha$ for which the points $(\alpha,2 + \alpha)$ and $\left( \frac{3}{2}\alpha,\alpha^{2} \right)$ lie o...A quadrilateral is inscribed in a parabola y2 = 4ax then :...The locus of the middle points of chords of the parabola $y^{2} = 4x$, which are of constant length ' $2l$ ' is :...Let $y^{2} = 4ax$ be a parabola and $x^{2} + y^{2} + 2bx = 0$ be a circle. If parabola and circle touch each other exter...Three normals are drawn from the point $(c,0)$ to the curve $y^{2} = x$. One normal is $x$-axis and the other two normal...