TS EAMCET · Maths · Ellipse
The length of the chord of the ellipse \(\frac{x^2}{4}+y^2=1\) formed on the line \(y=x+1\) is
- A \(2 \sqrt{2}\)
- B \(\frac{4}{5} \sqrt{2}\)
- C \(4 \sqrt{2}\)
- D \(\frac{8}{5} \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{8}{5} \sqrt{2}\)
Step-by-step Solution
Detailed explanation
Substitute \(y=x+1\) into \(\frac{x^2}{4}+y^2=1\): \(\frac{x^2}{4}+(x+1)^2=1\) \(x^2+4(x^2+2x+1)=4\) \(x^2+4x^2+8x+4=4\) \(5x^2+8x=0\) \(x(5x+8)=0\) \(x_1=0, x_2=-\frac{8}{5}\) Find corresponding y-coordinates: \(y_1=0+1=1\) \(y_2=-\frac{8}{5}+1=-\frac{3}{5}\) The intersection…
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