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  • The length of the straight line joining the feet of the perpendiculars let fall from the point <img alt="\mathrm{(p, q)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28p%2C%20

The length of the straight line joining the feet of the perpendiculars let fall from the point \mathrm{(p, q)}  on the pair of  lines\mathrm{ a x^2+2 h x y+b^2=0}  is   \mathrm{\frac{\sqrt{\left(p^2+q^2\right) 2} \sqrt{\left(h^2-a b\right)}}{\sqrt{\left\{(a-b)^2+k h^2\right\}}}} , where \mathrm{\mathrm{k}=}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Given pair of lines is \mathrm{a x^2+2 h x y+b y^2=0}..............(i)
Let \mathrm{ \theta}  be the angle between the lines represented by (i) and A, B be the feet of the perpendiculars from \mathrm{ \mathrm{P}(\mathrm{p}, \mathrm{q})}  on these lines.
Now if a circle be drawn with \mathrm{OP} as diameter, it will pass through both \mathrm{A}  and \mathrm{B} because angle in a semi-circle is a right angle.
\mathrm{\therefore \angle \mathrm{OAB}=\angle \mathrm{OPB}=\mathrm{\alpha}} (say) [angle in the same segment \mathrm{\mathrm{OB}} ]

\mathrm{\text { From } \triangle O A B \text {, we have } \frac{O B}{\sin \alpha}=\frac{A B}{\sin \theta}} .......(ii)

\mathrm{From \triangle \mathrm{OPB}, we have \frac{\mathrm{OP}}{\sin 90^{\circ}}=\frac{\mathrm{OB}}{\sin \alpha}=\frac{\mathrm{AB}}{\sin \theta} }    from (ii)...............(iii)

 If  \theta   be the acute angle between the pair of lines (i) 

then \mathrm{\tan \theta=\frac{2 \sqrt{\left(h^2-a b\right)}}{|a+b|}}   \mathrm{\sin \theta=}

\mathrm{\frac{2 \sqrt{\left(h^2-a b\right)}}{\sqrt{(a+b)^2+4 h^2-4 a b}}=\frac{2 \sqrt{\left(h^2-a b\right)}}{\sqrt{(a+b)^2+4 h^2}}}


 Now from (iii)    \mathrm{ A B=O P \sin \theta=\frac{\sqrt{\left(p^2+q^2\right) 2} \sqrt{\left(h^2-a b\right)}}{\sqrt{\left\{(a-b)^2+4 h^2\right\}}} }.

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Riya

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