Show that the two parabolas $x^{2} + 4 a (y - 2 b - a) = 0$ and $y^{2} = 4 b (x - 2 a + b)$ intersect at right angles at a common end of the latus rectum of each.

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Solution

$\frac{d y}{d x} = y^{'}$

$x^{2} + 4 a (y - 2 b - a) = 0$ ...............(1)

diffrentiate the eqn.

$2 x + 4 a y^{'} = 0$
$y^{'} = m = \frac{- x}{2 a}$

$y^{2} = 4 b (x - 2 a + b)$ ..........(2)

diffrentiate the eqn.

$2 y y^{'} = 4 b$

$y^{'} = n = \frac{2 b}{y}$

solving eqn (1) & (2)
$(2 a, 2 b)$

let $θ$ be the angle at intersection point

$m \times n = \frac{- x .2 b}{y .2 a} = \frac{- 2 a .2 b}{2 b .2 a} = - 1$

$\Rightarrow θ = 90^{\circ}$

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