Show that :
$(4 x + 5 y)^{2} - (4 x - 5 y)^{2} = 80 x y$

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Solution

$L . H . S$
$= (4 x + 5 y)^{2} - (4 x - 5 y)^{2}$
$= (4 x)^{2} + (5 y)^{2} + 2 (4 x) (5 y) - [(4 x)^{2} + (5 y)^{2} - 2 (4 x) (5 y)]$
$= 16 x^{2} + 25 y^{2} + 40 x y - [16 x^{2} + 25 y^{2} - 40 x y]$
$= 16 x^{2} + 25 y^{2} + 40 x y - 16 x^{2} - 25 y^{2} + 40 x y$
$= 40 x y + 40 x y$
$= 80 x y$
$= R . H . S$
Hence proved.

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Similar questions

Q. Tangents are drawn from the points on the parabola

y^{2} = - 8 (x + 4)

to the parabola

y^{2} = 4 x .

Then the locus of mid-point of chord of contact of

y^{2} = 4 x

Which of the following identities is not applicable to simplify
$(4 x + 5 y)^{2}$ - $(5 x + 4 y)^{2}$ ?

Q. The co – ordinates of the extremities of the latus rectum of the parabola

5 y^{2} = 4 x

are

The co-ordinates of the extremities of the latus rectum of the parabola $5 y^{2} = 4 x$ are

Solve:
(i) $\frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{4 x - 5}{2 x + 3}$
(ii) $\frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{y + 8}{1 + 2 y}$

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Show that :(4x+5y)2−(4x−5y)2=80xy

L.H.S=(4x+5y)2−(4x−5y)2=(4x)2+(5y)2+2(4x)(5y)−[(4x)2+(5y)2−2(4x)(5y)]=16x2+25y2+40xy−[16x2+25y2−40xy]=16x2+25y2+40xy−16x2−25y2+40xy=40xy+40xy=80xy=R.H.SHence proved.

Show that :
$(4 x + 5 y)^{2} - (4 x - 5 y)^{2} = 80 x y$