Let for two distinct values of p the lines y = x + p touch the ellipse at the points A and B. Let the line y = x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to : [2025]

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Solution

Q.
Let for two distinct values of p the lines y = x + p touch the ellipse $E : \frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$ at the points A and B. Let the line y = x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to : [2025]

1 24

2 36

3 48

4 20

Ans.
(1)

We have ellipse $E : \frac{x^{2}}{4^{2}} + \frac{y^{2}}{3^{2}} = 1$

and line y = x + p $\Rightarrow$ slope, m = 1

E and line y = x + p has point of contacts as A and B.

So, the point of contact

$= (\frac{\mp a^{2} m}{\sqrt{a^{2} m^{2} + b^{2}}}, \frac{\pm b^{2}}{\sqrt{a^{2} m^{2} + b^{2}}}) = (\frac{\mp 16}{5}, \frac{\pm 9}{5})$

Then, $A (\frac{- 16}{5}, \frac{9}{5}) and B (\frac{16}{5}, \frac{- 9}{5})$

Now, line y = x intersects with ellipse E at

$D (\frac{- 12}{5}, \frac{- 12}{5}) and C (\frac{12}{5}, \frac{12}{5})$

ABCD does not form any quadrilateral but if we do not consider the order then we have,

Area of $△ A B C = \frac{1}{2}$ $| \begin{matrix} - \frac{16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & - \frac{9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1 \end{matrix} |$ $= 12$

$∴$ Area of quadrilateral ABCD = 2 (Area of $△$ ABC) = 24 sq. units

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