TS EAMCET · Maths · Circle
The slopes of the focal chords of the parabola \(y^2=32 x\), which are tangents to the circle \(x^2+y^2=4\), are
- A \(\frac{1}{2}, \frac{-1}{2}\)
- B \(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\)
- C \(\frac{1}{\sqrt{15}}, \frac{-1}{\sqrt{15}}\)
- D \(\frac{2}{\sqrt{5}}, \frac{-2}{\sqrt{5}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{15}}, \frac{-1}{\sqrt{15}}\)
Step-by-step Solution
Detailed explanation
Given equation of circle is \[ x^2+y^2=(2)^2 \] \(\therefore\) The equation of tangent to the circle is \[ y=m x \pm 2 \sqrt{1+m^2}\left(\because y=m x \pm r \sqrt{1+m^2}\right) \] Also, equation of parabola is \[ y^2=32 x \] \(\therefore\) Focus of parabola is \((8,0)\). Since,…
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