Show that the distance between the points of intersection of the straight line x cos- - y sin- - p - 0 with the straight lines ax2 - 2hxy - by2 - 0 is 2p -h2 - abb cos2 - - 2h cos-sin- - a sin2 - -Deduce the area of the triangle formed by them-

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General Solution of cos theta = cos alpha

Show that the...

Question

Show that the distance between the points of intersection of the straight line $x cos α + y sin α - p = 0$ with the straight lines $a x^{2} + 2 h x y + b y^{2} = 0$ is $\frac{2 p \sqrt{h^{2} - a b}}{b {cos}^{2} α - 2 h cos α sin α + a {sin}^{2} α} .$
Deduce the area of the triangle formed by them.

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Solution

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

Let the lines represented by the given equation be $y = m_{1} x . . . (i)$ and $y = m_{2} x . . . . (i i)$
$m_{1} + m_{2} = \frac{- 2 h}{b} m_{1} m_{2} = \frac{a}{b}$

Given line $x cos α + y sin α = p . . . . . . (i i i)$

$A$ be the point of intersection $(i)$ and $(i i i)$

Solving the equations we get $x = \frac{p}{cos α + m_{1} sin α}, y = \frac{m_{1} p}{cos α + m_{1} sin α}$

$∴ A (\frac{p}{cos α + m_{1} sin α}, \frac{m_{1} p}{cos α + m_{1} sin α})$

Solving $(i i)$ and $(i i i)$

we get $x = \frac{p}{cos α + m_{2} sin α}, y = \frac{m_{2} p}{cos α + m_{2} sin α}$

$∴ B (\frac{p}{cos α + m_{2} sin α}, \frac{m_{2} p}{cos α + m_{2} sin α})$

Area of triangle $△ O A B$

$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 \frac{p}{cos α + m_{1} sin α} & \frac{m_{1} p}{cos α + m_{1} sin α} & 1 \frac{p}{cos α + m_{2} sin α} & \frac{m_{2} p}{cos α + m_{2} sin α} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = \frac{1}{2} \frac{p^{2} (m_{2} - m_{1})}{(cos α + m_{1} sin α) (cos α + m_{2} sin α)} = \frac{p^{2}}{2} \frac{\sqrt{{(m_{2} + m_{1})}^{2} - 4 m_{1} m_{2}}}{{cos}^{2} α + (m_{2} + m_{1}) cos α sin α + m_{1} m_{2} {sin}^{2} α} = \frac{p^{2}}{2} . \frac{\sqrt{\frac{{4 h}^{2}}{b^{2}} - \frac{4 a}{b}}}{{cos}^{2} α + (\frac{- 2 h}{b}) cos α sin α + \frac{a}{b} {sin}^{2} α} = \frac{p^{2} \sqrt{h^{2} - a b}}{b {cos}^{2} α - 2 h cos α sin α + a {sin}^{2} α}$

Also area of triangle $△ O A B$ $= \frac{1}{2} (O M) (A B)$

$O M = \frac{| 0. cos α + 0. sin α - p |}{\sqrt{{cos}^{2} α + {sin}^{2} α}} = p$

$\frac{1}{2} p (A B) = \frac{p^{2} \sqrt{h^{2} - a b}}{b {cos}^{2} α - 2 h cos α sin α + a {sin}^{2} α} \Rightarrow A B = \frac{2 p \sqrt{h^{2} - a b}}{b {cos}^{2} α - 2 h cos α sin α + a {sin}^{2} α}$

So the distance between the point of intersection is $\frac{2 p \sqrt{h^{2} - a b}}{b {cos}^{2} α - 2 h cos α sin α + a {sin}^{2} α}$

Hence proved.

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Show that the distance between the points of intersection of the straight line xcosα+ysinα−p=0 with the straight lines ax2+2hxy+by2=0 is 2p√h2−abbcos2α−2hcosαsinα+asin2α.Deduce the area of the triangle formed by them.