Question

Line $\frac{x}{a}+\frac{y}{b}=1$ cuts the coordinate axes at A (a, 0) and B(0, b) and the line $\frac{x}{a^{\prime }}+\frac{y}{b^{\prime }}=-1$ at A' (-a', 0) and B' (0, -b'). If the point A, B, A', B' are concyclic then the orthocenter of the triangle ABA' is

• A. (0, 0)
• B. (0, b')
• C. $(0,\frac{a{a}^{\prime }}{b})$
• D. $(0,\frac{b{b}^{\prime }}{a})$

Answer 1

Answer: (B), (C)

The correct options are
B

(0, b')

C

$(0,\frac{a{a}^{\prime }}{b})$

A, B, A', B' lies on circle.

$OA\times O{A}^{\prime }=OB\times O{B}^{\prime }a\times (-a^{\prime })=\left(b\right)(-b^{\prime })\left[\text{Power of a circle}\right]$

aa' = bb' ;

Equation of altitude through A' is y - 0 = $\left(\frac{a}{b}\right)$ (x + a');

If intersects the altitude x = 0 at $y=\frac{a{a}^{\prime }}{b}$ ;

Orthocenter are $(0,\frac{a{a}^{\prime }}{b})\&(0,b^{\prime })$