WBJEE · Maths · Hyperbola
Let \(P\) be the foot of the perpendicular from focus \(S\) of hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) on the line \(b x-a y=0\) and let \(C\) be the centre of the hyperbola. Then, the area of the rectangle whose sides are equal to that of \(S P\) and \(C P\) is
- A \(2 a b\)
- B \(a b\)
- C \(\frac{\left(a^{2}+b^{2}\right)}{2}\)
- D \(\frac{a}{b}\)
Answer & Solution
Correct Answer
(B) \(a b\)
Step-by-step Solution
Detailed explanation
Given, equation of hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) From figure, \(\begin{aligned} S P &=\mid \frac{a b e}{\sqrt{b^{2}+a^{2}}} \\ &=\left|\frac{a b e}{a e}\right|=b \end{aligned}\) and \(C S=a e\) Again, \(\Delta S P C\) is right angled triangle at…
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