Applications of Derivatives Question - None

Question

The maximum value of the sum of the intercepts made by any tangent to the curve (a sin² θ, 2a sin θ) with the axes is

Accepted Answer

Solution:

The curve is given by $x = a\sin^2 heta$, $y = 2a\sin heta$

The equation of any tangent to the curve is: $\frac{x}{a\sin heta} + \frac{y}{2a\sin^2 heta} = 1$

This gives the intercepts:

x-intercept $= a\sin heta$

y-intercept $= 2a\sin^2 heta$

Sum of intercepts $= a\sin^2 heta + 2a\sin heta = a\left(\sin^2 heta + 2\sin heta ight)$

$= a\left[\left(\sin heta + 1 ight)^2 - 1 ight]$

This is maximum when $\sin heta = 1$

Therefore, (Sum of intercepts)$_{max} = a(1 + 2) = 2a$

Hence, (a) is the correct answer.

Question: