MHT CET · Maths · Parabola
The intercept of the latusrectum to the parabola \(y^{2}=4 a x\) are \(b\) and \(k\), then \(k\) is equal to
- A \(\frac{a b}{a-b}\)
- B \(\frac{a}{b-a}\)
- C \(\frac{b}{b-a}\)
- D \(\frac{a b}{b-a}\)
Answer & Solution
Correct Answer
(D) \(\frac{a b}{b-a}\)
Step-by-step Solution
Detailed explanation
For latusrectum \(P Q\).
\(
y\left(t_{1}+t_{2}\right)-2 x-2 a t_{1} t_{2}=0
\)
and \(\quad t_{1} t_{2}=-1\)
For any point \(P(x, y)\), the focal distance is \(a+x\). \(\therefore \quad b=a+x=a+a t_{1}^{2}=a\left(1+t_{1}^{2}\right) \quad \ldots(\mathrm{i})\)
\(c=a+x=a+a t_{2}^{2}=a+\frac{a}{t_{1}^{2}} \quad\left(\because t_{1} t_{2}=-1\right)\)
\(=\frac{a\left(1+t_{1}^{2}\right)}{t_{1}^{2}}\)...(ii)
\(\frac{b}{c}=t_{1}^{2}\)
\(\therefore\) From Eq. (i), \(b=a\left(1+\frac{b}{c}\right)\)
\(\Rightarrow \quad b=a+\frac{a b}{c}\)
\(c=\frac{a b}{b-a}\)
\(
y\left(t_{1}+t_{2}\right)-2 x-2 a t_{1} t_{2}=0
\)
and \(\quad t_{1} t_{2}=-1\)
For any point \(P(x, y)\), the focal distance is \(a+x\). \(\therefore \quad b=a+x=a+a t_{1}^{2}=a\left(1+t_{1}^{2}\right) \quad \ldots(\mathrm{i})\)
\(c=a+x=a+a t_{2}^{2}=a+\frac{a}{t_{1}^{2}} \quad\left(\because t_{1} t_{2}=-1\right)\)
\(=\frac{a\left(1+t_{1}^{2}\right)}{t_{1}^{2}}\)...(ii)
\(\frac{b}{c}=t_{1}^{2}\)
\(\therefore\) From Eq. (i), \(b=a\left(1+\frac{b}{c}\right)\)
\(\Rightarrow \quad b=a+\frac{a b}{c}\)
\(c=\frac{a b}{b-a}\)
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