AP EAMCET · Maths · Hyperbola
The locus of a point whose chord of contact w.r.t. the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) touches the circle described on the straight line joining the foci of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) as diameter is
- A \(\frac{x^2}{a^4}-\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}\)
- B \(\frac{x^2}{a^4}-\frac{y^2}{b^4}=\frac{1}{\left(a^2-b^2\right)}\)
- C \(\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2-b^2\right)}\)
- D \(\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}\)
Answer & Solution
Correct Answer
(D) \(\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}\)
Step-by-step Solution
Detailed explanation
Circle on the join of foci \((a e, 0)\) and \((-a e, 0)\) diameter is: \(\begin{aligned} & \Rightarrow(x-a e)(x+a e)+(y-0)(y-0)=0 \\ & \Rightarrow x^2+y^2=a^2 e^2=a^2+b^2 \text { WC...(1) } \\ & {\left[a^2 e^2=a^2+b^2\right]}\end{aligned}\) Let chord of contact of…
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