Straight LineHardBloom L3
Question
The straight line, $ax + by = 1$ makes with the curve $px^{2} + 2axy + qy^{2} = r$ a chord which subtends a right angle at the origin. Then:
Options
A.$r\left( a^{2} + b^{2} \right) = p + q$
B.$r\left( a^{2} + p^{2} \right) = q + b$
C.$r\left( b^{2} + q^{2} \right) = p + a$
D.none of these
Solution
Sol. [Key Concept: A chord subtending a right angle at $O$: homogenise the conic with the line ⇒ coeff $x^{2}$ + coeff $y^{2 = 0}$.]
Homogeneous equation of the curve with line.
Coefficient of $x^{2} +$ coefficient of $y^{2} = 0$
homogenize the curve with the chord $ax + by = 1$:
$${px}^{2 + 2}axy + {qy}^{2 =}r(ax + by)^{2}$$
$$\Rightarrow \ (p - {ra}^{2)}x^{2 + 2(}a - rab)xy + (q - {rb}^{2)}y^{2 = 0}$$
This pair $OA,OB$ subtends a right angle at $O$
⇒ coeff $x^{2}$ + coeff $y^{2 = 0}$:
$(p - {ra}^{2) + (}q - {rb}^{2) = 0} \Rightarrow \ r(a^{2 +}b^{2) =}p + q$ → (a).
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