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  • The line bisects the two distinct chords to the ellipse <img alt="\mathrm{\frac{x^

The line \mathrm{x+y=b} bisects the two distinct chords to the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=2} which pass through the point \mathrm{P(a,-b)} on it. Then

Option: 1

\mathrm{a^2+7 a b \geq 6 b^2}


Option: 2

\mathrm{a^2+6 a b \geq 7 b^2}


Option: 3

\mathrm{a^2+4 a b \geq 3 b^2}


Option: 4

\mathrm{a^2+2 a b \geq 6 b^2}


Answers (1)

best_answer

Choose the mid-point R on the line \mathrm{x+y=b} as \mathrm{(\lambda, b-\lambda).}

If Q be \mathrm{\left(x_1, y_1\right)} then R is mid-point of PQ

\mathrm{ \therefore \quad 2 \lambda=a+x_1 \text { and } 2(b-\lambda)=-b+y_1 }

\mathrm{ \therefore \quad x_1=2 \lambda-a, y_1=3 b-2 \lambda }
Now \mathrm{\left(x_1, y_1\right)} lies on ellipse

\mathrm{ \therefore \frac{(2 \lambda-a)^2}{a^2}+\frac{(3 b-2 \lambda)^2}{b^2}=2 }

or \mathrm{b^2\left(4 \lambda^2-4 \lambda a+a^2\right)+a^2\left(9 b^2-12 b \lambda+4 \lambda^2\right)-2 a^2 b^2=0}

Arrange as a quadratic in variable \mathrm{\lambda}

\mathrm{4 \lambda^2\left(b^2+a^2\right)-4 \lambda a b(b+3 a)+8 a^2 b^2=0}

or \mathrm{\lambda^2\left(b^2+a^2\right)-\lambda a b(b+3 a)+2 a^2 b^2=0}

Above is a quadratic in variable \mathrm{\lambda}. Since two distinct chords are drawn, we must have two real roots of the above

quadratic. Hence its discriminant \mathrm{\Delta \geq 0.}

\mathrm{ \begin{aligned} & \therefore \quad a^2 b^2(b+3 a)^2-4 \cdot 2 a^2 b^2\left(b^2+a^2\right) \geq 0 \\\\ & \text { or } b^2+9 a^2+6 a b-\left(8 b^2+8 a^2\right) \geq 0 \end{aligned} }

or \mathrm{a^2+6 a b \geq 7 b^2} is the required condition.

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shivangi.bhatnagar

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