Solve the problem.
-Write an equation for the tangent line to the curve $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 100 } = 1$ at the point $\left( \frac { 7 } { \sqrt { 2 } } , \frac { 10 } { \sqrt { 2 } } \right)$.
A) $\frac { x } { 10 } + \frac { y } { 7 } = 1$
B) $\frac { x } { 10 } + \frac { y } { 7 } = \sqrt { 2 }$
C) $\frac { x } { 7 } + \frac { y } { 10 } = 1$
D) $\frac { x } { 7 } + \frac { y } { 10 } = \sqrt { 2 }$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the given ellipse at the specified point, we first differentiate the equation of the ellipse implicitly with respect to $x$, which gives us $\frac{2x}{49} + \frac{2yy'}{100} = 0$. Solving for $y'$ (the slope of the tangent) at the point $\left( \frac{7}{\sqrt{2}}, \frac{10}{\sqrt{2}} \right)$, we get $y' = -\frac{x}{y} \cdot \frac{100}{49}$. Substituting $x = \frac{7}{\sqrt{2}}$ and $y = \frac{10}{\sqrt{2}}$, we find the slope of the tangent line. Then, using the point-slope form of a line, we can derive the equation of the tangent. The correct equation that matches this process is $\frac{x}{7} + \frac{y}{10} = \sqrt{2}$, which corresponds to choice D.

Solve the Problem x249+y2100=1\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 100 } = 149x2​+100y2​=1

Solve the Problem $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 100 } = 1$