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Lines and Angles

A rod of leng...

Question

A rod of length ‘l’ slides with its ends on two perpendicular lines. Then the locus of its midpoint is

x^{2} + y^{2} = \frac{l^{2}}{4}

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x^{2} + y^{2} = \frac{l^{2}}{2}

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x^{2} - y^{2} = \frac{l^{2}}{4}

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x^{2} - y^{2} = \frac{l^{2}}{3}

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Solution

The correct option is A $x^{2} + y^{2} = \frac{l^{2}}{4}$

The correct answer is option (a). Check the video for the approach.

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x^{2} + y^{2} + 2 x - 4 = 0

x^{2} + y^{2} - y + 1 = 0

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If the curves $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and $\frac{x^{2}}{l^{2}} - \frac{y^{2}}{m^{2}} = 1$ cut each other orthogonally, then

y^{2} = 4 a (x - l_{1})

x^{2} = 4 a (y - l_{2})

l_{1}

l_{2}

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