Question

A variable line is drawn through

A (α, β)

meeting the line-pair

a x^{2} + 2 h x y + b y^{2} = 0

, in

P

and

Q

. A point

R

is taken on the line such that

A R

is the harmonic mean between

A P

and

A Q

. Prove that the locus of

R

is

(a α + h β) x + (h α + b β) y = 0

.

Accepted Answer

Let the lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ be $O P$ is

$y = m_{1} x$

and $O Q$ is

$y = m_{2} x$ where $m_{1} + m_{2} = - \frac{2 h}{b} and m_{1} m_{2} = \frac{a}{b}$

Equation of variable line passing through $A (α, β)$ and having slope $m$ is

$y - β = m (x - α) . . . . (i)$

For point $P$ , put $y = m_{1} x in (i)$

$m_{1} x - β = m (x - α)$

$\Rightarrow (m_{1} - m) x = β - m α$

$∴ x = \frac{β - m α}{m_{1} - m}$

and $y = \frac{m_{1} (β - m α)}{m_{1} - m}$

$∴$ Co-ordinate of $P = (\frac{β - m α}{m_{1} - m}, \frac{m_{1} (β - m α)}{m_{1} - m})$

Similarly, co-ordinate of $Q = (\frac{β - m α}{m_{2} - m}, \frac{m_{2} (β - m α)}{m_{2} - m})$

$∴ A P = \sqrt{{(\frac{β - m α}{m_{1} - m} - α)}^{2} + {(\frac{m_{1} (β - m α)}{m_{1} - m} - β)}^{2}}$

$= \sqrt{{(\frac{β - m_{1} α}{m_{1} - m})}^{2} + {(\frac{m β - m m_{1} α}{m_{1} - m})}^{2}}$

$= \sqrt{\frac{β^{2} + {m_{1}}^{2} α^{2} - 2 m_{1} α β + m^{2} β^{2} - 2 m^{2} m_{1} α β + m^{2} {m_{1}}^{2} α^{2}}{{(m_{1} - m)}^{2}}}$

$∴ A P = \sqrt{\frac{(1 + m^{2}) (β^{2} + {m_{1}}^{2} α^{2} 2 m_{1} α β)}{{(m_{1} - m)}^{2}}}$

Similarly,

$A Q = \sqrt{\frac{(1 + m^{2}) (β^{2} + {m_{2}}^{2} α^{2} - 2 m_{2} α β)}{{(m_{2} - m)}^{2}}}$

Let co-ordinate of $R be (p, q)$ which is on the line $(i)$

$∴ m = \frac{q - β}{p - α}$

Also, given that $A R$ is harmonic mean of $A P & A Q$

$∴ A R = \frac{2 . A P . A Q}{A P + A Q}$

$\Rightarrow \sqrt{{(q - β)}^{2} + {(p - α)}^{2}} = \frac{2 (1 + m^{2}) \sqrt{\frac{(β^{2} + {m_{1}}^{2} α^{2} - 2 m_{1} α β)}{{(m_{1} - m)}^{2}}} . \sqrt{\frac{(β^{2} + {m_{2}}^{2} α^{2} - 2 m_{2} α β)}{{(m_{2} - m)}^{2}}}}{\sqrt{\frac{(1 + m^{2}) (β^{2} + {m_{1}}^{2} α^{2} - 2 m_{1} α β)}{{(m_{1} - m)}^{2}}} + \sqrt{\frac{(1 + m^{2}) (β^{2} + m_{2} α^{2} - 2 m_{2} α β)}{{(m_{2} - m)}^{2}}}}$

Squaring both sides, we have

${(q - β)}^{2} + {(p - α)}^{2} = \frac{4 {(1 + m^{2})}^{2} [\frac{β^{2} + {m_{1}}^{2} α^{2} - 2 m_{1} α β}{{(m_{1} - m)}^{2}}] [\frac{β^{2} + {m_{2}}^{2} α^{2} - 2 m_{2} α β}{{(m_{2} - m)}^{2}}]}{\frac{(1 + m^{2}) (β^{2} + {m_{1}}^{2} α^{2} - 2 m_{1} α β)}{{(m_{1} - m)}^{2}} + \frac{(1 + m^{2}) (β^{2} + {m_{2}}^{2} α^{2} - 2 m_{2} α β)}{{(m_{2} - m)}^{2}} + 2 \frac{(1 + m^{2})}{(m_{1} - m) (m_{2} - m)} \sqrt{(β^{2} + {m_{1}}^{2} α^{2} - 2 m_{1} α β) (β^{2} + {m_{1}}^{2} α^{2} - 2 m_{2} α β)}}$

Now, substituting the values of $m_{1} + m_{2} = - \frac{2 h}{a}, m_{1} m_{2} = \frac{a}{b} & m = \frac{q - β}{p - α}$ , we get

$(a α + h β) p + (h α + b β) q = 0$

Replacing $p, q by x, y$ , we get

$(a α + h β) x + (h α + b β) y = 0$ , which is the required locus.

Dr. SK Goyal Solutions for Chapter: Pair of Straight Lines, Exercise 7: EXERCISE ON LEVEL-II

Dr. SK Goyal Mathematics Solutions for Exercise - Dr. SK Goyal Solutions for Chapter: Pair of Straight Lines, Exercise 7: EXERCISE ON LEVEL-II

Questions from Dr. SK Goyal Solutions for Chapter: Pair of Straight Lines, Exercise 7: EXERCISE ON LEVEL-II with Hints & Solutions

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