The length of the chord of the ellipse x²/25 + y²/16 = 1, whose mid point is ≤ft(1, 2/5), is equal to : [27-Jan-2024 Shift 1]

Correct Answer is : 16915{{1691}}{5}51691

Correct Answer is : 16915{{1691}}{5}51691

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Section: Mathematics

Question : 41 of 117

Marks: +1, -0

\frac{x^2}{25} + \frac{y^2}{16} = 1

\left(1, \frac{2}{5}\right)

Solution: 👈: Video Solution

Equation of chord with given middle point.

T = S_1

\frac{x}{25} + \frac{y}{40} = \frac{1}{25} + \frac{1}{100}

\frac{8x+5y}{200} = \frac{8+2}{200}

y = \frac{10-8x}{5}

........(i)

\frac{x^2}{25} + \frac{(10-8x)^2}{400} = 1 \quad \text{(put in original equation)}

\frac{16x^2 + 100 + 64x^2 - 160x}{400} = 1

4x^2 - 8x - 15 = 0

x = \frac{8 \pm \sqrt{304}}{8}

x_1 = \frac{8 + \sqrt{304}}{8} ; \quad x_2 = \frac{8 - \sqrt{304}}{8}

\text{Similarly, } y = \frac{10 - 18 \pm \sqrt{304}}{5}

= \frac{2 \pm \sqrt{304}}{5}

y_1 = \frac{2 - \sqrt{304}}{5} ; \quad y_2 = \frac{2 + \sqrt{304}}{5}

\text{Distance } = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

= \sqrt{ \frac{4 \times 304}{64} + \frac{4 \times 304}{25} } = \frac{\sqrt{1691}}{5}

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