MHT CET · Maths · Hyperbola
The normal to the rectangular hyperbola \(x y=c^{2}\) at the point ' ' ' meets the curve again at a point \({ }^{\prime} t^{\prime \prime}\), such that
- A \(t^{2} t^{\prime}=-1\)
- B \(t^{3} t^{\prime}=-1\)
- C \(t t^{\prime}=-1\)
- D None of these
Answer & Solution
Correct Answer
(B) \(t^{3} t^{\prime}=-1\)
Step-by-step Solution
Detailed explanation
Equation of normal at \((c t, c / c t)\) is \(y t-x t^{3}-c+c t^{4}=0\)
If it passes through any point say \(Q\) having its coordinates as ' \(t\) ' i.e., \(\left(\mathrm{ct}^{\prime} \mathrm{c} / \mathrm{t}^{\prime}\right)\)
The coordinates must satisfy the equation. Hence, we get \(\frac{c}{t^{\prime}} \cdot t-c t^{\prime} t^{3}-c+c t^{4}=0\)
\(
\Rightarrow \quad t-t^{\prime 2} t^{3}-t^{\prime}+t^{\prime} t^{4}=0
\)
Factorizing, we get, \(\left(t^{\prime} t^{3}+1\right)\left(t-t^{\prime}\right)=0\)
\(
\text { Hencet }^{\prime} t^{3}+1=0 \text { or } t^{\prime} t^{3}=-1
\)
as \(t \neq t^{\prime}\)
If it passes through any point say \(Q\) having its coordinates as ' \(t\) ' i.e., \(\left(\mathrm{ct}^{\prime} \mathrm{c} / \mathrm{t}^{\prime}\right)\)
The coordinates must satisfy the equation. Hence, we get \(\frac{c}{t^{\prime}} \cdot t-c t^{\prime} t^{3}-c+c t^{4}=0\)
\(
\Rightarrow \quad t-t^{\prime 2} t^{3}-t^{\prime}+t^{\prime} t^{4}=0
\)
Factorizing, we get, \(\left(t^{\prime} t^{3}+1\right)\left(t-t^{\prime}\right)=0\)
\(
\text { Hencet }^{\prime} t^{3}+1=0 \text { or } t^{\prime} t^{3}=-1
\)
as \(t \neq t^{\prime}\)
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