The length of the chord intercepted by the ellipse $x^{2} + 2 y^{2} = 4$ on the normal at the point $(\sqrt{2}, 1)$ is:

\frac{3 \sqrt{6}}{5}

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\frac{2 \sqrt{6}}{5}

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\frac{7 \sqrt{6}}{5}

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\frac{6 \sqrt{6}}{5}

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Solution

The correct option is C $\frac{6 \sqrt{6}}{5}$
Equation of ellipse is $x^{2} + 2 y^{2} = 4$ .....(1)
or $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$
Here, $a^{2} = 4, b^{2} = 2$
Equation of normal to ellipse at $(\sqrt{2}, 1)$ is
$\frac{4 x}{\sqrt{2}} - \frac{2 y}{1} = 4 - 2$
or $y = \sqrt{2} x - 1$ ....(2)
Solving (1) and (2) simultaneously, we get
$5 x^{2} - 4 \sqrt{2} x - 2 = 0$
$\Rightarrow x_{1} + x_{2} = \frac{4 \sqrt{2}}{5}, x_{1} x_{2} = \frac{- 2}{5}$
$(x_{1} - x_{2})^{2} = (x_{1} + x_{2})^{2} - 4 x_{1} x_{2} = \frac{16 \times 2}{25} + \frac{8}{5} = \frac{72}{25}$
Using $(2)$ , $(y_{1} - y_{2})^{2} = 2 (x_{1} - x_{2})^{2} = \frac{2 \times 72}{25}$
The length of the chord is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
$= \sqrt{\frac{3 \times 72}{25}} = \frac{6 \sqrt{6}}{5}$

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Question 6

Which of the following expressions show that rational numbers are associative under multiplication.
(a) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = (\frac{2}{3} \times \frac{- 6}{7}) \times \frac{3}{5}$
(b) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = \frac{2}{3} \times (\frac{3}{5} \times \frac{- 6}{7})$
(c) $\frac{2}{3} \times (\frac{- 6}{7} \times \frac{3}{5}) = (\frac{3}{5} \times \frac{2}{3}) \times \frac{- 6}{7}$
(d) $(\frac{2}{3} \times \frac{- 6}{7}) \times \frac{3}{5} = (\frac{- 6}{7} \times \frac{2}{3}) \times \frac{3}{5}$

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The length of the chord intercepted by the ellipse x2+2y2=4 on the normal at the point (√2,1) is:

The length of the chord intercepted by the ellipse $x^{2} + 2 y^{2} = 4$ on the normal at the point $(\sqrt{2}, 1)$ is: