First slide

Equation of line in Straight lines

Question

$The locus of point of intersection of the line y + m x = \sqrt{a^{2} m^{2} + b^{2}} and m y - x = \sqrt{a^{2} + b^{2} m^{2}} is$

Moderate

A

$x^{2} + y^{2} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$
B

$x^{2} + y^{2} = a^{2} + b^{2}$
C

$x^{2} - y^{2} = a^{2} - b^{2}$
D

$\frac{1}{x^{2}} + \frac{1}{y^{2}} = a^{2} - b^{2}$

Solution

Let the point of intersection of given two lines is P(h, k),
which lies on both the lines.
$∴ k + m h = \sqrt{a^{2} m^{2} + b^{2}} and m k - h = \sqrt{a^{2} + b^{2} m^{2}}$
Squaring and adding, we get
${(k + m h)}^{2} + {(m k - h)}^{2} = (a^{2} m^{2} + b^{2}) (a^{2} + b^{2} m^{2})$
$\begin{array}{l} (1 + m^{2}) k^{2} + (1 + m^{2}) h^{2} = a^{2} m^{2} + b^{2} + a^{2} + b^{2} m^{2} \\ (1 + m^{2}) (k^{2} + h^{2}) = a^{2} (m^{2} + 1) + b^{2} (m^{2} + 1) \\ ∴ locus of (h, k) is x^{2} + y^{2} = a^{2} + b^{2} \end{array}$

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