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10-2. Parabola, Ellipse, Hyperbola
normal
Let $L$ is distance between two parallel normals of , $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,\,\,a > b$ then maximum value of $L$ is
A
$2a$
B
$2b$
C
$a+b$
D
$2(a -b)$
Solution
$\operatorname{axsec} \theta-\operatorname{bycosec} \theta=a^{2}-b^{2}$
$\operatorname{axsec} \theta-$ $bycosec$ $\theta=-\left(a^{2}-b^{2}\right)$
$L = \frac{{2\left( {{a^2} – {b^2}} \right)}}{{\sqrt {{a^2}{{\sec }^2}\theta + {b^2}\cos e{c^2}\theta } }}$
$\because $ $\quad {a^2}{\sec ^2}\theta + {b^2}\cos e{c^2}\theta \ge {(a + b)^2}$
$\mathrm{L} \leq 2(\mathrm{a}-\mathrm{b})$
Standard 11
Mathematics
