QJEEM23Ma8abd: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8abd Solution: The set of all values of for which the line bisects two…

Question

Verified Solution • 2025–26 Session moderate

The set of all values of

a^{2}

for which the line

x+y=0

bisects two distinct chords drawn from a point

\mathrm{P}\left(\frac{1+a}{2}, \frac{1-a}{2}\right)

on the circle

2 x^{2}+2 y^{2}-(1+a) x-(1-a) y=0

, is equal to :

Accepted Answer

x^{2}+y^{2}-\frac{(1+a) x}{2}-\frac{(1-a) y}{2}=0

Centre

\left(\frac{1+\mathrm{a}}{4}, \frac{1-\mathrm{a}}{4}\right) \Rightarrow(\mathrm{h}, \mathrm{k})

\mathrm{P}\left(\frac{1+\mathrm{a}}{2}, \frac{1-\mathrm{a}}{2}\right) \Rightarrow(2 \mathrm{h}, 2 \mathrm{k})

Equation of chord

\Rightarrow \mathrm{T}=\mathrm{S}_{1}

\Rightarrow(\mathrm{x}-\mathrm{y}) \lambda-\frac{2 \mathrm{~h}(\mathrm{x}+\lambda)}{2}-\frac{(2 \mathrm{k})(\mathrm{y}-\lambda)}{2}

=2 \lambda^{2}-2 \mathrm{~h}(\lambda)+2 \mathrm{k} \lambda

Now,

\lambda(2 \mathrm{~h}, 2 \mathrm{k})

satisfies the chord

\therefore(2 \mathrm{~h}-2 \mathrm{k}) \lambda-\mathrm{h}(\mathrm{x}+\lambda)-\mathrm{k}(\mathrm{y}-\lambda)

\Rightarrow 2 \lambda^{2}+4 \mathrm{k} \lambda-4 \mathrm{~h} \lambda+\mathrm{h} \lambda-\mathrm{k} \lambda+\mathrm{hx}+\mathrm{ky}=0

\Rightarrow 2 \lambda^{2}+\lambda(3 \mathrm{k}-3 \mathrm{~h})+\mathrm{ky}+\mathrm{hx}=0

\Rightarrow \mathrm{D}&gt;0

\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8(\mathrm{ky}+\mathrm{hx})&gt;0

\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8\left(2 \mathrm{k}^{2}+2 \mathrm{~h}^{2}\right)&gt;0

\Rightarrow-7 \mathrm{k}^{2}-7 \mathrm{~h}^{2}-18 \mathrm{kh}&gt;0

\Rightarrow 7 \mathrm{k}^{2}+7 \mathrm{~h}^{2}+18 \mathrm{kh}&lt;0

\Rightarrow 7\left(\frac{1-\mathrm{a}}{4}\right)^{2}+7\left(\frac{1+\mathrm{a}}{4}\right)^{2}+18\left(\frac{1-\mathrm{a}^{2}}{16}\right)&lt;0

\Rightarrow 7\left[\frac{2\left(1+\mathrm{a}^{2}\right)}{16}\right]+\frac{18\left(1-\mathrm{a}^{2}\right)}{16}&lt;0, \quad \mathrm{a}^{2}=\mathrm{t}

\Rightarrow \frac{7}{8}(1+\mathrm{t})+\frac{18(1-\mathrm{t})}{16}&lt;0

\Rightarrow \frac{14+14 \mathrm{t}+18-18 \mathrm{t}}{16}&lt;0

\Rightarrow 4 \mathrm{t}&gt;32

\Rightarrow

\mathrm{t}&gt;8

\Rightarrow

\mathrm{a}^{2}&gt;8

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