Let a tangent to the curve $9 x^2+16 y^2=144$ intersect the coordinate axes at the points $A$ and $B$. Then, the minimum length of the line segment $AB$ is
The equation of the ellipse is:
\[
\frac{x^2}{16} + \frac{y^2}{9} = 1.
\]
The equation of the tangent at point \( P(4\cos\theta, 3\sin\theta) \) is:
\[
\frac{x \cos\theta}{4} + \frac{y \sin\theta}{3} = 1.
\]
The tangent intersects the \( x \)-axis at \( A(4\sec\theta, 0) \) and the \( y \)-axis at \( B(0, 3\csc\theta) \).
The length of \( AB \) is:
\[
AB = \sqrt{(4\sec\theta)^2 + (3\csc\theta)^2}.
\]
Simplify:
\[
AB = \sqrt{16\sec^2\theta + 9\csc^2\theta}.
\]
Using the identity:
\[
\sec^2\theta + \csc^2\theta \geq 2,
\]
we find:
\[
AB \geq \sqrt{25 + 16\tan^2\theta + 9\cot^2\theta} \geq 7.
\]
The minimum length of \( AB \) is:
\[
\boxed{7}.
\]
A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
Given a function y = f(x), the equation of the tangent for this curve at x = x0
Slope of tangent (at x=x0) m=dy/dx||x=x0
A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
The normal to a given curve y = f(x) at a point x = x0
The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
