To find the length of the chord of the ellipse whose equation is given by \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) and with the mid-point \(\left(1, \frac{2}{5}\right)\), we will use the chord length formula:
The equation of a chord with midpoint \((\alpha, \beta)\) for an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is given by:
\(T = S_1\),
where \(T\) is the transformed equation of the chord and \(S_1\) represents the expression obtained by replacing \() by \(\alp\) and \() by \(\be\) in the equation of ellipse.
For the given ellipse, \(a^2 = 25\) and \(b^2 = 16\).
Plugging \((\alpha, \beta) = \left(1, \frac{2}{5}\right)\), the equation \(S_1\) becomes:\)
\(\frac{1^2}{25} + \frac{\left(\frac{2}{5}\right)^2}{16} = 1\).
Calculate the components:
Therefore, \(\frac{1}{25} + \frac{1}{100} = 0.04 + 0.01 = 0.05\).
Now, the length of the chord \(L\) is given by:
\(L = \sqrt{\frac{4c^2}{a^2b^2 - (b^2\cos^2\theta + a^2\sin^2\theta)}}\)
Here, \(c\) is calculated as:
\(c = \sqrt{a^2b^2(1 - (\text{sum from }S_1))} = \sqrt{25 \cdot 16 \cdot 0.95} = \sqrt{380}\).
The expression simplifies to the final length formula with:
\(L = \frac{\sqrt{1691}}{5}\),
aligning with the correct answer \(\frac{\sqrt{1691}}{5}\).
Hence, the length of the chord is indeed \(\frac{\sqrt{1691}}{5}\).
Given the ellipse:
\(\frac{x^2}{25} + \frac{y^2}{16} = 1\) and a chord with midpoint \(\left( 1, \frac{25}{8} \right)\).
Step 1. Equation of the Chord: The chord equation is:
\(\frac{x}{25} + \frac{y}{40} = 1 \Rightarrow y = \frac{200 - 8x}{5}\)
Step 2. Substitute into the Ellipse: Substituting \( y \) gives:
\(\frac{x^2}{25} + \frac{\left( \frac{200 - 8x}{5} \right)^2}{16} = 1\)
Simplifying:
\(2x^2 - 80x + 990 = 0 \Rightarrow x = 20 \pm \sqrt{10}\)
Step 3. Length of the Chord: Using the distance formula, the length is:
\(\text{Length} = \frac{\sqrt{1691}}{5}\)
In a △ABC, suppose y = x is the equation of the bisector of the angle B and the equation of the side AC is 2x−y = 2. If 2AB = BC and the points A and B are respectively (4, 6) and (α, β), then α + 2β is equal to:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,