A chord ax-y-1 subtends 90- at the centre of the circle x2-y2-32- Then the value of a is

By homogenization, x^2+y^2=32(ax+y)^2 ⇒ 2x^2+2y^2=3(ax+y)^2 ⇒ (3a^2 2)x^2+y^2+6axy=0 This pair of straight lines subtends 90^∘ at the origin. ⇒Coefficient of ...

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Standard XII

Mathematics

Homogenization

A chord ax+y=...

Question

A chord $a x + y = 1$ subtends $90^{\circ}$ at the centre of the circle $x^{2} + y^{2} = \frac{3}{2} .$ Then the value of $a$ is

\pm \frac{1}{\sqrt{2}}

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\pm \frac{1}{\sqrt{3}}

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\pm \sqrt{2}

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\pm \sqrt{3}

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Solution

The correct option is B $\pm \frac{1}{\sqrt{3}}$

By homogenization,
$x^{2} + y^{2} = \frac{3}{2} (a x + y)^{2} \Rightarrow 2 x^{2} + 2 y^{2} = 3 (a x + y)^{2} \Rightarrow (3 a^{2} - 2) x^{2} + y^{2} + 6 a x y = 0$

This pair of straight lines subtends $90^{\circ}$ at the origin.
$\Rightarrow Coefficient of x^{2} + Coefficient of y^{2} = 0 \Rightarrow 3 a^{2} - 1 = 0 \Rightarrow a = \pm \frac{1}{\sqrt{3}}$

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A chord ax+y=1 subtends 90∘ at the centre of the circle x2+y2=32. Then the value of a is

The correct option is B ±1√3 By homogenization, x2+y2=32(ax+y)2⇒2x2+2y2=3(ax+y)2⇒(3a2−2)x2+y2+6axy=0 This pair of straight lines subtends 90∘ at the origin. ⇒Coefficient of x2+Coefficient of y2=0⇒3a2−1=0⇒a=±1√3

A chord $a x + y = 1$ subtends $90^{\circ}$ at the centre of the circle $x^{2} + y^{2} = \frac{3}{2} .$ Then the value of $a$ is