The line given by the equation $y - \frac { 5 \sqrt { 3 } } { 2 } = \frac { 4 \sqrt { 3 } } { 5 } ( x - 2 )$ is perpendicular to the tangent line to the ellipse $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 25 } = 1$ at the point x = 2.

Question

Quizplus · Accepted Answer

The slope of the given line is $\frac{4\sqrt{3}}{5}$. For the ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1$, the slope of the tangent line at any point can be found using implicit differentiation. Differentiating both sides with respect to $x$ gives $\frac{x}{8} + \frac{yy'}{25} = 0$, where $y'$ is the derivative of $y$ with respect to $x$ (i.e., the slope of the tangent line). At $x = 2$, the corresponding $y$ value on the ellipse can be found by substituting $x = 2$ into the ellipse equation, yielding $y = \pm 5\sqrt{3}/2$. Substituting $x = 2$ and $y = 5\sqrt{3}/2$ into the differentiated equation gives a slope of $-\frac{5}{4\sqrt{3}}$ for the tangent line at that point. Since the product of the slopes of two perpendicular lines is $-1$, and $\frac{4\sqrt{3}}{5} \cdot -\frac{5}{4\sqrt{3}} = -1$, the given line is indeed perpendicular to the tangent line to the ellipse at $x = 2$.

The Line Given by the Equation y−532=435(x−2)y - \frac { 5 \sqrt { 3 } } { 2 } = \frac { 4 \sqrt { 3 } } { 5 } ( x - 2 )y−253​​=543​​(x−2)

The Line Given by the Equation $y - \frac { 5 \sqrt { 3 } } { 2 } = \frac { 4 \sqrt { 3 } } { 5 } ( x - 2 )$