A circle centre = (-15, 0) and radius = 15/2 Chord to circle through (-30, 0) & tangent to y2 = 30x. Length the of chord = ?
(a) \(\frac{15}{\sqrt{5}}\)
(b) \(\frac{17}{\sqrt{7}}\)
(c) \(\frac{19}{\sqrt{5}}\)
(d) \(\frac{10}{\sqrt{3}}\)
Correct answer: (a) \(\frac{15}{\sqrt{5}}\)
Explanation:
y2 = 30x
4a = 30
⟹ a = \(\frac{15}{2}\)
Equation of tangent
y = mx + \(\frac{a}{m}\)
⟹ y = mx + \(\frac{15}{2m}\)
Passes through (-30, 0)
⟹ 0 = -30m + \(\frac{15}{2m}\)
⟹ 30 m = \(\frac{15}{2m}\)
⟹ m2 = \(\frac{1}{4}\)
⟹ m = \(\pm \frac{1}{2}\)
y = \(\frac{x}{2} + 15\)
Distance of line from centre is
\(\frac{|\frac{-15}{2} + 15|}{\sqrt{1^2 + \frac{1}{2^2}}}\)
= \(\frac{\frac{15}{2}}{\frac{\sqrt 5}{2}}\)
= \(\frac{15}{\sqrt 5}\)
= y = \(\frac{-x}{2}-15\)
Distance of line from centre is
\(\frac{|\frac{15}{2} - 15|}{\sqrt{1^2 + \frac{1}{1^2}}}\) = \(\frac{15}{\sqrt 5}\)
Length of chord = \(2\sqrt{(\frac{15}{2})^2 -(\frac{15}{\sqrt5})^2}\)
= \(2 \times 15 \sqrt{\frac{1}{4} - \frac{1}{5}}\)
= \(30 \sqrt{\frac{1}{20}}\) = \(\frac{30}{\sqrt {20}}\)
= \(\frac{15}{\sqrt{5}}\)