Question
Any tangent at a point $P(x, y)$ to the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{18}=1$ meets the co-ordinate axes in the points $A$ and $B$ such that the area of the triangle $\mathrm{OAB}$ is least, then find the point $P$.
Step 1
The equation of the tangent line at point \(P(x_0, y_0)\) can be derived using the formula: \[ \frac{x_0 x}{8} + \frac{y_0 y}{18} = 1 \] Show more…
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Any tangent at a point $P(x, y)$ to the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{18}=1$ meets the co-ordinate axes in the points $A$ and $B$ such that the area of the triangle $O A B$ is least, then the point $P$ is (a) $(\sqrt{8}, 0)$ (b) $(0, \sqrt{18})$ (c) $(2,3)$ (d) None
The Tangent and Normal
Level II
Consider the tangent line to the ellipse x^2/a^2 + y^2/b^2 = 1 at a point (p,q) in the first quadrant. a) Find the x- intercept and y-intercept of the tangent line. b) Find the minimum length of the portion of the tangent line cut off by the coordinate axes. c) Find the minimum area of the triangle formed by the tangent line and the corrdinate axes.
Find the equation of the line that is tangent to the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ in the first quadrant and forms with the coordinate axes the triangle with smallest possible area ( $a$ and $b$ are positive constants).
Applications of the Derivative
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